Ok_ch15-Deep Foundation I-pile Foundation

CHAPTER 15DEEP FOUNDATION I:

PILE FOUNDATION

15.1 INTRODUCTIONShallow foundations are normally used where the soil close to the ground surface and up to thezone of significant stress possesses sufficient bearing strength to carry the superstructure loadwithout causing distress to the superstructure due to settlement. However, where the top soil iseither loose or soft or of a swelling type the load from the structure has to be transferred to deeperfirm strata.

The structural loads may be transferred to deeper firm strata by means of piles. Piles are longslender columns either driven, bored or cast-in-situ. Driven piles are made of a variety of materialssuch as concrete, steel, timber etc., whereas cast-in-situ piles are concrete piles. They may besubjected to vertical or lateral loads or a combination of vertical and lateral loads. If the diameter ofa bored-cast-in-situ pile is greater than about 0.75 m, it is sometimes called a drilled pier, drilledcaisson or drilled shaft. The distinction made between a small diameter bored cast-in-situ pile (lessthan 0.75 m) and a larger one is just for the sake of design considerations. The design of drilledpiers is dealt with in Chapter 17. This chapter is concerned with driven piles and small diameterbored cast-in-situ piles only.

15.2 CLASSIFICATION OF PILESPiles may be classified as long or short in accordance with the Lid ratio of the pile (whereL = length, d = diameter of pile). A short pile behaves as a rigid body and rotates as a unit underlateral loads. The load transferred to the tip of the pile bears a significant proportion of the totalvertical load on the top. In the case of a long pile, the length beyond a particular depth loses itssignificance under lateral loads, but when subjected to vertical load, the frictional load on the sidesof the pile bears a significant part to the total load.

605

606 Chapter 15

Piles may further be classified as vertical piles or inclined piles. Vertical piles are normallyused to carry mainly vertical loads and very little lateral load. When piles are inclined at an angleto the vertical, they are called batter piles or raker piles. Batter piles are quite effective for takinglateral loads, but when used in groups, they also can take vertical loads. The behavior of verticaland batter piles subjected to lateral loads is dealt with in Chapter 16.

Types of Piles According to Their CompositionPiles may be classified according to their composition as

1. Timber Piles,

2. Concrete Piles,

3. Steel Piles.

Timber Piles: Timber piles are made of tree trunks with the branches trimmed off. Such piles shallbe of sound quality and free of defects. The length of the pile may be 15 m or more. If greaterlengths are required, they may be spliced. The diameter of the piles at the butt end may vary from30 to 40 cm. The diameter at the tip end should not be less than 15 cm.

Piles entirely submerged in water last long without decay provided marine borers are notpresent. When a pile is subjected to alternate wetting and drying the useful life is relatively shortunless treated with a wood preservative, usually creosote at 250 kg per m3 for piles in fresh waterand 350 kg/m3 in sea water.

After being driven to final depth, all pile heads, treated or untreated, should be sawed squareto sound undamaged wood to receive the pile cap. But before concrete for the pile cap is poured, thehead of the treated piles should be protected by a zinc coat, lead paint or by wrapping the pile headswith fabric upon which hot pitch is applied.

Driving of timber piles usually results in the crushing of the fibers on the head (or brooming)which can be somewhat controlled by using a driving cap, or ring around the butt.

The usual maximum design load per pile does not exceed 250 kN. Timber piles are usuallyless expensive in places where timber is plentiful.

Concrete Piles. Concrete piles are either precast or cast-in-situ piles. Precast concrete pilesare cast and cured in a casting yard and then transported to the site of work for driving. If the workis of a very big nature, they may be cast at the site also.

Precast piles may be made of uniform sections with pointed tips. Tapered piles may bemanufactured when greater bearing resistance is required. Normally piles of square or octagonalsections are manufactured since these shapes are easy to cast in horizontal position. Necessaryreinforcement is provided to take care of handling stresses. Piles may also be prestressed.

Maximum load on a prestressed concrete pile is approximately 2000 kN and on precast piles1000 kN. The optimum load range is 400 to 600 kN.

Steel Piles. Steel piles are usually rolled H shapes or pipe piles, //-piles are proportioned towithstand large impact stresses during hard driving. Pipe piles are either welded or seamless steelpipes which may be driven either open-end or closed-end. Pipe piles are often filled with concreteafter driving, although in some cases this is not necessary. The optimum load range on steel piles is400 to 1200kN.

15.3 TYPES OF PILES ACCORDING TO THE METHOD OF INSTALLATION

According to the method of construction, there are three types of piles. They are

1. Driven piles,2. Cast-in-situ piles and

3. Driven and cast-in-situ piles.

Deep Foundation I: Pile Foundation 607

Driven Piles

Piles may be of timber, steel or concrete. When the piles are of concrete, they are to be precast.They may be driven either vertically or at an angle to the vertical. Piles are driven using a pilehammer. When a pile is driven into granular soil, the soil so displaced, equal to the volume of thedriven pile, compacts the soil around the sides since the displaced soil particles enter the soil spacesof the adjacent mass which leads to densification of the mass. The pile that compacts the soiladjacent to it is sometimes called a compaction pile. The compaction of the soil mass around a pileincreases its bearing capacity.

If a pile is driven into saturated silty or cohesive soil, the soil around the pile cannot bedensified because of its poor drainage qualities. The displaced soil particles cannot enter the voidspace unless the water in the pores is pushed out. The stresses developed in the soil mass adjacent tothe pile due to the driving of the pile have to be borne by the pore water only. This results in thedevelopment of pore water pressure and a consequent decrease in the bearing capacity of the soil.The soil adjacent to the piles is remolded and loses to a certain extent its structural strength. Theimmediate effect of driving a pile in a soil with poor drainage qualities is, therefore, to decrease itsbearing strength. However, with the passage of time, the remolded soil regains part of its loststrength due to the reorientation of the disturbed particles (which is termed thixotrophy} and due toconsolidation of the mass. The advantages and disadvantages of driven piles are:

Advantages

1. Piles can be precast to the required specifications.2. Piles of any size, length and shape can be made in advance and used at the site. As a result,

the progress of the work will be rapid.3. A pile driven into granular soil compacts the adjacent soil mass and as a result the bearing

capacity of the pile is increased.4. The work is neat and clean. The supervision of work at the site can be reduced to a

minimum. The storage space required is very much less.5. Driven piles may conveniently be used in places where it is advisable not to drill holes for

fear of meeting ground water under pressure.6. Drivens pile are the most favored for works over water such as piles in wharf structures or

jetties.

Disadvantages

1. Precast or prestressed concrete piles must be properly reinforced to withstand handlingstresses during transportation and driving.

2. Advance planning is required for handling and driving.3. Requires heavy equipment for handling and driving.4. Since the exact length required at the site cannot be determined in advance, the method

involves cutting off extra lengths or adding more lengths. This increases the cost of theproject.

5. Driven piles are not suitable in soils of poor drainage qualities. If the driving of piles is notproperly phased and arranged, there is every possibility of heaving of the soil or the liftingof the driven piles during the driving of a new pile.

6. Where the foundations of adjacent structures are likely to be affected due to the vibrationsgenerated by the driving of piles, driven piles should not be used.

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Cast-in-situ Piles

Cast-in-situ piles are concrete piles. These piles are distinguished from drilled piers as smalldiameter piles. They are constructed by making holes in the ground to the required depth and thenfilling the hole with concrete. Straight bored piles or piles with one or more bulbs at intervals maybe cast at the site. The latter type are called under-reamed piles. Reinforcement may be used as perthe requirements. Cast-in-situ piles have advantages as well as disadvantages.

Advantages

1. Piles of any size and length may be constructed at the site.

2. Damage due to driving and handling that is common in precast piles is eliminated in thiscase.

3. These piles are ideally suited in places where vibrations of any type are required to beavoided to preserve the safety of the adjoining structure.

4. They are suitable in soils of poor drainage qualities since cast-in-situ piles do notsignificantly disturb the surrounding soil.

Disadvantages

1. Installation of cast-in-situ piles requires careful supervision and quality control of all thematerials used in the construction.

2. The method is quite cumbersome. It needs sufficient storage space for all the materialsused in the construction.

3. The advantage of increased bearing capacity due to compaction in granular soil that couldbe obtained by a driven pile is not produced by a cast-in-situ pile.

4. Construction of piles in holes where there is heavy current of ground water flow or artesianpressure is very difficult.

A straight bored pile is shown in Fig. 15.1 (a).

Driven and Cast-in-situ Piles

This type has the advantages and disadvantages of both the driven and the cast-in-situ piles. Theprocedure of installing a driven and cast-in-situ pile is as follows:

A steel shell is driven into the ground with the aid of a mandrel inserted into the shell. Themandrel is withdrawn and concrete is placed in the shell. The shell is made of corrugated andreinforced thin sheet steel (mono-tube piles) or pipes (Armco welded pipes or common seamlesspipes). The piles of this type are called a shell type. The shell-less type is formed by withdrawing theshell while the concrete is being placed. In both the types of piles the bottom of the shell is closed witha conical tip which can be separated from the shell. By driving the concrete out of the shell an enlargedbulb may be formed in both the types of piles. Franki piles are of this type. The common types ofdriven and cast-in-situ piles are given in Fig. 15.1. In some cases the shell will be left in place and thetube is concreted. This type of pile is very much used in piling over water.

15.4 USES OF PILESThe major uses of piles are:

1. To carry vertical compression load.

2. To resist uplift load.

3. To resist horizontal or inclined loads.


Fluted steelshell filledwith concrete

Franki pile\

Corrugatedsteel shellfilled with \concrete

Bulb

(a) (b) (c) (d)

Figure 15.1 Types of cast-in-situ and driven cast-in-situ concrete piles

Normally vertical piles are used to carry vertical compression loads coming fromsuperstructures such as buildings, bridges etc. The piles are used in groups joined together by pilecaps. The loads carried by the piles are transferred to the adjacent soil. If all the loads coming on thetops of piles are transferred to the tips, such piles are called end-bearing or point-bearing piles.However, if all the load is transferred to the soil along the length of the pile such piles are calledfriction piles. If, in the course of driving a pile into granular soils, the soil around the pile getscompacted, such piles are called compaction piles. Fig. 15.2(a) shows piles used for the foundationof a multistoried building to carry loads from the superstructure.

Piles are also used to resist uplift loads. Piles used for this purpose are called tension piles oruplift piles or anchor piles. Uplift loads are developed due to hydrostatic pressure or overturningmovement as shown in Fig. 15.2(a).

Piles are also used to resist horizontal or inclined forces. Batter piles are normally used toresist large horizontal loads. Fig. 15.2(b) shows the use of piles to resist lateral loads.

15.5 SELECTION OF PILEThe selection of the type, length and capacity is usually made from estimation based on the soilconditions and the magnitude of the load. In large cities, where the soil conditions are well knownand where a large number of pile foundations have been constructed, the experience gained in thepast is extremely useful. Generally the foundation design is made on the preliminary estimatedvalues. Before the actual construction begins, pile load tests must be conducted to verify the designvalues. The foundation design must be revised according to the test results. The factors that governthe selection of piles are:

1. Length of pile in relation to the load and type of soil2. Character of structure3. Availability of materials4. Type of loading5. Factors causing deterioration

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A multi-storied building on piles

Uplift oranchor piles

^ Upliftpile

-^CCompressionpile

Piles used to resist uplift loads

Figure 15.2(a) Principles of floating foundation; and a typical rigid raft foundation

Retainingwall Bridge pier

v/JW\>!*K

Batter pile

Figure 15.2(b) Piles used to resist lateral loads

6. Ease of maintenance7. Estimated costs of types of piles, taking into account the initial cost, life expectancy and

cost of maintenance8. Availability of funds

All the above factors have to be largely analyzed before deciding up on a particular type.

15.6 INSTALLATION OF PILESThe method of installing a pile at a site depends upon the type of pile. The equipment required for thispurpose varies. The following types of piles are normally considered for the purpose of installation

1. Driven piles

The piles that come under this category are,

a. Timber piles,


b. Steel piles, //-section and pipe piles,c. Precast concrete or prestressed concrete piles, either solid or hollow sections.

2. Driven cast-in-situ piles

This involves driving of a steel tube to the required depth with the end closed by a detachableconical tip. The tube is next concreted and the shell is simultaneously withdrawn. In some cases theshell will not be withdrawn.

3. Bored cast-in-situ piles

Boring is done either by auguring or by percussion drilling. After boring is completed, the bore isconcreted with or without reinforcement.

Pile Driving Equipment for Driven and Driven Cast-in-situ PilesPile driving equipment contains three parts. They are

1. A pile frame,2. Piling winch,3. Impact hammers.

Pile Frame

Pile driving equipment is required for driven piles or driven cast-in-situ piles. The driving pileframe must be such that it can be mounted on a standard tracked crane base machine for mobility onland sites or on framed bases for mounting on stagings or pontoons in offshore construction. Fig.15.3 gives a typical pile frame for both onshore and offshore construction. Both the types must becapable of full rotation and backward or forward raking. All types of frames consist essentially ofleaders, which are a pair of steel members extending for the full height of the frame and whichguide the hammer and pile as it is driven into the ground. Where long piles have to be driven theleaders can be extended at the top by a telescopic boom.

The base frame may be mounted on swivel wheels fitted with self-contained jacking screwsfor leveling the frame or it may be carried on steel rollers. The rollers run on steel girders or longtimbers and the frame is moved along by winching from a deadman set on the roller track, or byturning the rollers by a tommy-bar placed in holes at the ends of the rollers. Movements parallel tothe rollers are achieved by winding in a wire rope terminating in hooks on the ends of rollers; theframe then skids in either direction along the rollers. It is important to ensure that the pile frameremains in its correct position throughout the driving of a pile.

Piling Winches

Piling winches are mounted on the base. Winches may be powered by steam, diesel or gasolineengines, or electric motors. Steam-powered winches are commonly used where steam is used forthe piling hammer. Diesel or gasoline engines, or electric motors (rarely) are used in conjunctionwith drop hammers or where compressed air is used to operate the hammers.

Impact Hammers

The impact energy for driving piles may be obtained by any one of the following types of hammers.They are

1. Drop hammers,2. Single-acting steam hammers,3. Double-acting steam hammers,

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To crane for lifting.I Assembly rests on pileA during driving

Staticr weight

Oscillator u i_r-

Pile

9310mm

(a) The Ackermanns M14-5P pile frame (b) Diagrammatic sketch of vibratory pile driver

Figure 15.3 Pile driving equipment and vibratory pile driver

4. Diesel hammer,5. Vibratory hammer.

Drop hammers are at present used for small jobs. The weight is raised and allowed to fallfreely on the top of the pile. The impact drives the pile into the ground.

In the case of a single-acting steam hammer steam or air raises the moveable part of thehammer which then drops by gravity alone. The blows in this case are much more rapidly deliveredthan for a drop hammer. The weights of hammers vary from about 1500 to 10,000 kg with thelength of stroke being about 90 cm. In general the ratio of ram weight to pile weight may vary from0.5 to 1.0.

In the case of a double-acting hammer steam or air is used to raise the moveable part of thehammer and also to impart additional energy during the down stroke. The downward accelerationof the ram owing to gravity is increased by the acceleration due to steam pressure. The weights ofhammers vary from about 350 to 2500 kg. The length of stroke varies from about 20 to 90 cm. Therate of driving ranges from 300 blows per minute for the light types, to 100 blows per minute for theheaviest types.

Diesel or internal combustion hammers utilize diesel-fuel explosions to provide the impactenergy to the pile. Diesel hammers have considerable advantage over steam hammers because theyare lighter, more mobile and use a smaller amount of fuel. The weight of the hammer varies fromabout 1000 to 2500 kg.


The advantage of the power-hammer type of driving is that the blows fall in rapid succession(50 to 150 blows per minute) keeping the pile in continuous motion. Since the pile is continuouslymoving, the effects of the blows tend to convert to pressure rather than impact, thus reducingdamage to the pile.

The vibration method of driving piles is now coming into prominence. Driving is quiet anddoes not generate local vibrations. Vibration driving utilizes a variable speed oscillator attached tothe top of the pile (Fig. 15.3(b)). It consists of two counter rotating eccentric weights which are inphase twice per cycle (180° apart) in the vertical direction. This introduces vibration through thepile which can be made to coincide with the resonant frequency of the pile. As a result, a push-pulleffect is created at the pile tip which breaks up the soil structure allowing easy pile penetration intothe ground with a relatively small driving effort. Pile driving by the vibration method is quitecommon in Russia.

Jetting Piles

Water jetting may be used to aid the penetration of a pile into dense sand or dense sandy gravel.Jetting is ineffective in firm to stiff clays or any soil containing much coarse to stiff cobbles orboulders.

Where jetting is required for pile penetration a stream of water is discharged near the pilepoint or along the sides of the pile through a pipe 5 to 7.5 cm in diameter. An adequate quantity ofwater is essential for jetting. Suitable quantities of water for jetting a 250 to 350 mm pile are

Fine sand 15-25 liters/second,

Coarse sand 25-40 liters/second,

Sandy gravels 45-600 liters/second.

A pressure of at least 5 kg/cm2 or more is required.

PART A—VERTICAL LOAD BEARING CAPACITY OF ASINGLE VERTICAL PILE

15.7 GENERAL CONSIDERATIONSThe bearing capacity of groups of piles subjected to vertical or vertical and lateral loads dependsupon the behavior of a single pile. The bearing capacity of a single pile depends upon

1. Type, size and length of pile,2. Type of soil,3. The method of installation.

The bearing capacity depends primarily on the method of installation and the type of soilencountered. The bearing capacity of a single pile increases with an increase in the size and length.The position of the water table also affects the bearing capacity.

In order to be able to design a safe and economical pile foundation, we have to analyze theinteractions between the pile and the soil, establish the modes of failure and estimate thesettlements from soil deformation under dead load, service load etc. The design should complywith the following requirements.

1. It should ensure adequate safety against failure; the factor of safety used depends on theimportance of the structure and on the reliability of the soil parameters and the loadingsystems used in the design.

614 Chapter 15

2. The settlements should be compatible with adequate behavior of the superstructure toavoid impairing its efficiency.

Load Transfer Mechanism

Statement of the Problem

Fig. 15.4(a) gives a single pile of uniform diameter d (circular or any other shape) and length Ldriven into a homogeneous mass of soil of known physical properties. A static vertical load isapplied on the top. It is required to determine the ultimate bearing capacity Qu of the pile.

When the ultimate load applied on the top of the pile is Qu, a part of the load is transmitted tothe soil along the length of the pile and the balance is transmitted to the pile base. The loadtransmitted to the soil along the length of the pile is called the ultimate friction load or skin load Qf

and that transmitted to the base is called the base or point load Qb. The total ultimate load Qu isexpressed as the sum of these two, that is,

Qu = Qb + Qf=qbAb+fsAs (15.1)

where Qu = ultimate load applied on the top of the pileqb = ultimate unit bearing capacity of the pile at the baseAb = bearing area of the base of the pileAs = total surface area of pile embedded below ground surface

fs - unit skin friction (ultimate)

Load Transfer Mechanism

Consider the pile shown in Fig. 15.4(b) is loaded to failure by gradually increasing the load on thetop. If settlement of the top of the pile is measured at every stage of loading after an equilibriumcondition is attained, a load settlement curve as shown in Fig. 15.4(c) can be obtained.

If the pile is instrumented, the load distribution along the pile can be determined at differentstages of loading and plotted as shown in Fig. 15.4(b).

When a load Q{ acts on the pile head, the axial load at ground level is also Qr but at level Al

(Fig. 15.4(b)), the axial load is zero. The total load Q{ is distributed as friction load within a lengthof pile L{. The lower section A{B of pile will not be affected by this load. As the load at the top isincreased to Q2, the axial load at the bottom of the pile is just zero. The total load Q2 is distributedas friction load along the whole length of pile L. The friction load distribution curves along the pileshaft may be as shown in the figure. If the load put on the pile is greater than <22, a part of this loadis transferred to the soil at the base as point load and the rest is transferred to the soil surroundingthe pile. With the increase of load Q on the top, both the friction and point loads continue toincrease. The friction load attains an ultimate value £J,at a particular load level, say Qm, at the top,and any further increment of load added to Qm will not increase the value of Q~ However, the pointload, Q , still goes on increasing till the soil fails by punching shear failure. It has been determinedby Van Wheele (1957) that the point load Q increases linearly with the elastic compression of thesoil at the base.

The relative proportions of the loads carried by skin friction and base resistance depend onthe shear strength and elasticity of the soil. Generally the vertical movement of the pile which isrequired to mobilize full end resistance is much greater than that required to mobilize full skinfriction. Experience indicates that in bored cast-in-situ piles full frictional load is normallymobilized at a settlement equal to 0.5 to 1 percent of pile diameter and the full base load Qb at 10 to20 percent of the diameter. But, if this ultimate load criterion is applied to piles of large diameter inclay, the settlement at the working load (with a factor of safety of 2 on the ultimate load) may beexcessive. A typical load-settlement relationship of friction load and base load is shown in


Curves showing theloads carried by

pile shaft

_ [ _ ( _ , Base or Tip of pile

Q.

(b) Load-transfer curves

Load (kN)„() 1000 2000 3000 4000 5000

25

I5 0

<U

2 75*\ * ~*

100

125

150

Shaftload

Baseload

(c) Load-settlement curve (d) Load-settlement relationships forlarge-diameter bored and cast-in-place piles(after Tomlinson, 1986)

Figure 15.4 Load transfer mechanism

Fig. 15.4(d) (Tomlinson, 1986) for a large diameter bored and cast-in-situ pile in clay. It may beseen from this figure that the full shaft resistance is mobilized at a settlement of only 15 mmwhereas the full base resistance, and the ultimate resistance of the entire pile, is mobilized at asettlement of 120 mm. The shaft load at a settlement of 15 mm is only 1000 kN which is about 25percent of the base resistance. If a working load of 2000 kN at a settlement of 15 mm is used for thedesign, at this working load, the full shaft resistance will have been mobilized whereas only about50 percent of the base resistance has been mobilized. This means if piles are designed to carry aworking load equal to 1/3 to 1/2 the total failure load, there is every likelihood of the shaftresistance being fully mobilized at the working load. This has an important bearing on the design.

The type of load-settlement curve for a pile depends on the relative strength values of thesurrounding and underlying soil. Fig. 15.5 gives the types of failure (Kezdi, 1975). They are as follows:

616 Chapter 1 5

(c)

S = Settlementrs = Shear strengthQ = load on the pile (e)

Figure 15.5 Types of failure of piles. Figures (a) to (e) indicate how strength of soildetermines the type of failure: (a) buckling in very weak surrounding soil; (b) general

shear failure in the strong lower soil; (c) soil of uniform strength; (d) low strengthsoil in the lower layer, skin friction predominant; (e) skin friction in tension

(Kezdi, 1975)

Fig 15.5(a) represents a driven- pile (wooden or reinforced concrete), whose tip bears on avery hard stratum (rock). The soil around the shaft is too weak to exert any confining pressure orlateral resistance. In such cases, the pile fails like a compressed, slender column of the samematerial; after a more or less elastic compression buckling occurs. The curve shows a definitefailure load.

Fig. 15.5(b) is the type normally met in practice. The pile penetrates through layers of soilhaving low shear strength down to a layer having a high strength and the layer extending sufficientlybelow the tip of the pile. At ultimate load Qu, there will be a base general shear failure at the tip of thepile, since the upper layer does not prevent the formation of a failure surface. The effect of the shaftfriction is rather less, since the lower dense layer prevents the occurrence of excessive settlements.Therefore, the degree of mobilization of shear stresses along the shaft will be low. The load settlementdiagram is of the shape typical for a shallow footing on dense soil.

Fig. 15.5(c) shows the case where the shear strength of the surrounding soil is fairly uniform;therefore, a punching failure is likely to occur. The load-settlement diagram does not have a verticaltangent, and there is no definite failure load. The load will be carried by point resistance as well asby skin friction.

Fig 15.5(d) is a rare case where the lower layer is weaker. In such cases, the load will be carriedmainly by shaft friction, and the point resistance is almost zero. The load-settlement curve shows avertical tangent, which represents the load when the shaft friction has been fully mobilized.


Fig. 15.5(e) is a case when a pull, -Q, acts on the pile. Since the point resistance is again zerothe same diagram, as in Fig. 15.5(d), will characterize the behavior, but heaving occurs.

Definition of Failure Load

The methods of determining failure loads based on load-settlement curves are described insubsequent sections. However, in the absence of a load settlement curve, a failure load may bedefined as that which causes a settlement equal to 10 percent of the pile diameter or width (as perthe suggestion of Terzaghi) which is widely accepted by engineers. However, if this criterion isapplied to piles of large diameter in clay and a nominal factor of safety of 2 is used to obtain theworking load, then the settlement at the working load may be excessive.

Factor of SafetyIn almost all cases where piles are acting as structural foundations, the allowable load is governedsolely from considerations of tolerable settlement at the working load.

The working load for all pile types in all types of soil may be taken as equal to the sum of thebase resistance and shaft friction divided by a suitable factor of safety. A safety factor of 2.5 isnormally used. Therefore we may write

Q -^^ (.5.2)a 2.5

In case where the values of Qb and Q.can be obtained independently, the allowable load canbe written as

Q =^- + - (15.3)a 3 1.5

It is permissible to take a safety factor equal to 1.5 for the skin friction because the peak valueof skin friction on a pile occurs at a settlement of only 3-8 mm (relatively independent of shaftdiameter and embedded length but may depend on soil parameters) whereas the base resistancerequires a greater settlement for full mobilization.

The least of the allowable loads given by Eqs. (15.2) and (15.3) is taken as the design workingload.

15.8 METHODS OF DETERMINING ULTIMATE LOAD BEARINGCAPACITY OF A SINGLE VERTICAL PILEThe ultimate bearing capacity, Qu, of a single vertical pile may be determined by any of thefollowing methods.

1. By the use of static bearing capacity equations.2. By the use of SPT and CPT values.3. By field load tests.4. By dynamic method.

The determination of the ultimate point bearing capacity, qb, of a deep foundation on the basisof theory is a very complex one since there are many factors which cannot be accounted for in thetheory. The theory assumes that the soil is homogeneous and isotropic which is normally not thecase. All the theoretical equations are obtained based on plane strain conditions. Only shape factorsare applied to take care of the three-dimensional nature of the problem. Compressibility

618 Chapter 15

characteristics of the soil complicate the problem further. Experience and judgment are thereforevery essential in applying any theory to a specific problem. The skin load Q, depends on the natureof the surface of the pile, the method of installation of the pile and the type of soil. An exactevaluation of QAs a difficult job even if the soil is homogeneous over the whole length of the pile.The problem becomes all the more complicated if the pile passes through soils of variablecharacteristics.

15.9 GENERAL THEORY FOR ULTIMATE BEARING CAPACITYAccording to Vesic (1967), only punching shear failure occurs in deep foundations irrespective ofthe density of the soil so long as the depth-width ratio Lid is greater than 4 where L = length of pileand d = diameter (or width of pile). The types of failure surfaces assumed by different investigatorsare shown in Fig. 15.6 for the general shear failure condition. The detailed experimental study ofVesic indicates that the failure surfaces do not revert back to the shaft as shown in Fig. 15.6(b).

The total failure load Qu may be written as follows

Q = n + W = Q, +Q, + W (154)*-•« x-'u p *-•£> *-•/ p \i~1-^}

where Qu = load at failure applied to the pileQb = base resistanceQs = shaft resistance

W = weight of the pile.The general equation for the base resistance may be written as

Qb= cNc+q'0Nq+-ydNY Ab (15.5)

where d — width or diameter of the shaft at base levelq' o = effective overburden pressure at the base level of the pileAb = base area of pile

c - cohesion of soily = effective unit weight of soil

Nc, N , N = bearing capacity factors which take into account the shape factor.

Cohesionless Soils

For cohesionless soils,term qoN for deep foundations. Therefore Eq. (15.5) reduces toFor cohesionless soils, c = 0 and the term l/2ydN becomes insignificant in comparison with the

(15.6)

Eq. (15.4) may now be written as

Qb-Qu+Wp=q'0NqAb + Wp+Qf (15.7)

The net ultimate load in excess of the overburden pressure load qoAb is

If we assume, for all practical purposes, W and q'0Ab are roughly equal for straight side ormoderately tapered piles, Eq. (15.8) reduces to

ars

Highlight

ars

Highlight

Deep Foundation I: Pile Foundation

Qu

619

\Qu

L 111 ill , I , , , ,_M46

ill

(a) (c)

(b)

Figure 15.6 The shapes of failure surfaces at the tips of piles as assumed by (a)Terzaghi, (b) Meyerhof, and (c) Vesic

or (15.9)

where As = surface area of the embedded length of the pileq'o = average effective overburden pressure over the embedded depth of the pileKs = average lateral earth pressure coefficient

8 = angle of wall friction.

Cohesive SoilsFor cohesive soils such as saturated clays (normally consolidated), we have for </> = 0, N - 1 and N= 0. The ultimate base load from Eq. (15.5) is

620

Qb=(chNc+q'o)Ab

The net ultimate base load is

Chapter 15

(15.10)

Therefore, the net ultimate load capacity of the pile, Q , is

or Qu=cbNcAb+Asa cu (15.12)

where a = adhesion factorcu = average undrained shear strength of clay along the shaftcb = undrained shear strength of clay at the base level

NC = bearing capacity factorEquations (15.9) and (15.12) are used for analyzing the net ultimate load capacity of piles in

cohesionless and cohesive soils respectively. In each case the following types of piles areconsidered.

1 . Driven piles2. Driven and cast-in-situ piles3. Bored piles

15.10 ULTIMATE BEARING CAPACITY IN COHESIONLESS SOILS

Effect of Pile Installation on the Value of the Angle of Friction

When a pile is driven into loose sand its density is increased (Meyerhof, 1959), and the horizontalextent of the compacted zone has a width of about 6 to 8 times the pile diameter. However, in densesand, pile driving decreases the relative density because of the dilatancy of the sand and theloosened sand along the shaft has a width of about 5 times the pile diameter (Kerisel, 1961). On thebasis of field and model test results, Kishida (1967) proposed that the angle of internal frictiondecreases linearly from a maximum value of 02 at the pile tip to a low value of 0t at a distance of3.5dfrom the tip where d is the diameter of the pile, 0j is the angle of friction before the installationof the pile and 02 after the installation as shown in Fig. 15.7. Based on the field data, the relationshipbetween 0l and 02 in sands may be written as

(j), +40(15.13)

Figure 15.7 The effect of driving a pile on


An angle of 0, = 02 = 40° in Eq. (15.13) means no change of relative density due to piledriving. Values of 0} are obtained from insitu penetration tests (with no correction due tooverburden pressure, but corrected for field procedure) by using the relationships establishedbetween 0 and SPT or CPT values. Kishida (1967) has suggested the following relationshipbetween 0 and the SPT value Ncor as

0° =^20Ncor+l5° (15.14)

However, Tomlinson (1986) is of the opinion that it is unwise to use higher values for 0 dueto pile driving. His argument is that the sand may not get compacted, as for example, when piles aredriven into loose sand, the resistance is so low and little compaction is given to the soil. He suggeststhat the value of 0 used for the design should represent the in situ condition that existed beforedriving.

With regard to driven and cast-in-situ piles, there is no suggestion by any investigator as towhat value of 0 should be used for calculating the base resistance. However, it is safer to assume theinsitu 0 value for computing the base resistance.

With regard to bored and cast-in-situ piles, the soil gets loosened during boring. Tomlinson(1986) suggests that the 0 value for calculating both the base and skin resistance should represent theloose state. However, Poulos et al., (1980) suggests that for bored piles, the value of 0be taken as

0=^ -3 (15.15)

where 0j = angle of internal friction prior to installation of the pile.

15.11 CRITICAL DEPTHThe ultimate bearing capacity Qu in cohesionless soils as per Eq. (15.9) is

Qu = 9'oNqAb+«'0Ks imSAs (15.16a)

or Qu=1b\+fA (15.16b)

Eq. (15.16b) implies that both the point resistance qb and the skin resistance fs are functionsof the effective overburden pressure qo in cohesionless soils and increase linearly with the depth ofembedment, L, of the pile. However, extensive research work carried out by Vesic (1967) hasrevealed that the base and frictional resistances remain almost constant beyond a certain depth ofembedment which is a function of 0. This phenomenon was attributed to arching by Vesic. Oneconclusion from the investigation of Vesic is that in cohesionless soils, the bearing capacity factor,N , is not a constant depending on 0 only, but also on the ratio Lid (where L = length of embedmentof pile, d = diameter or width of pile). In a similar way, the frictional resistance, fs, increases withthe L/d ratio and remains constant beyond a particular depth. Let LC be the depth, which may becalled the critical depth, beyond which both qb and/5 remain constant. Experiments of Vesic haveindicated that Lc is a function of 0. The LJd ratio as a function of 0 may be expressed as follows(Poulos and Davis, 1980)

For 28° <0< 36.5°

LJd = 5 + 0.24 (0° - 28°) (15.17a)

For 36.5° < 0 < 42°

LJd = 1 + 2.35(0° - 36.5°) (15.17b)

The above expressions have been developed based on the curve given by Poulos and Davis,(1980) giving the relationship between LJd and 0°.

ars

Highlight

622 Chapter 15

The Eqs. (15.17) indicate

LJd=5 at 0 = 28°

Lc/d = l at 0 = 36.5°

Lc/d=20 at 0 = 42°

The 0 values to be used for obtaining LJd are as follows (Poulos and Davis, 1980)

for driven piles 0 = 0.75 0j + 10° (15.18a)

for bored piles: 0 = 0, - 3° (15.18b)

where 0j = angle of internal friction prior to the installation of the pile.

15.12 TOMLINSON'S SOLUTION FOR Qb IN SANDDriven PilesThe theoretical N factor in Eq. (15.9) is a function of 0. There is great variation in the values ofNderived by different investigators as shown in Fig. 15.8. Comparison of observed base resistancesof piles by Nordlund (1963) and Vesic (1964) have shown (Tomlinson, 1986) that N valuesestablished by Berezantsev et al., (1961) which take into account the depth to width ratio of the pile,

CQ

1000

100

10

1. Terzaghi(1943)2. Vesic (1963)3. Berezantsev (1961)4. BrinchHansen(1951)5. Skemptonetal(1953)6. Caquot-Kerisel(1956)

I / //

77

1. BrinchHansen(1961)8. Meyerhof( 1953) Bored Piles9. Meyerhof( 1953) Driven Piles

10. De Beer (1945)

25 30 35 40 45Angle of internal friction 0°

50

Figure 15.8 Bearing capacity factors for circular deep foundations (after Kezdi, 1975)


200

623

150

£ 100

50

20 25 30 35 40Angle of internal friction 0°

45

Figure 15.9 Berezantsev's bearing capacity factor, N (after Tomlinson, 1986)

most nearly conform to practical criteria of pile failure. Berezantsev's values of N as adopted byTomlinson (1986) are given in Fig. 15.9.

It may be seen from Fig. 15.9 that there is a rapid increase in TV for high values of 0, givingthereby high values of base resistance. As a general rule (Tomlinson, 1986), the allowable workingload on an isolated pile driven to virtual refusal, using normal driving equipment, in a dense sand orgravel consisting predominantly of quartz particles, is given by the allowable load on the pileconsidered as a structural member rather than by consideration of failure of the supporting soil, orif the permissible working stress on the material of the pile is not exceeded, then the pile will notfail.

As per Tomlinson, the maximum base resistance qb is normally limited to 11000 kN/m2

(110 t/ft2) whatever might be the penetration depth of the pile.

Bored and Cast-in-situ Piles in Cohesionless SoilsBored piles are formed in cohesionless soils by drilling with rigs. The sides of the holes might besupported by the use of casing pipes. When casing is used, the concrete is placed in the drilled holeand the casing is gradually withdrawn. In all the cases the sides and bottom if the hole will beloosened as a result of the boring operations, even though it may be initially be in a dense ormedium dense state. Tomlinson suggests that the values of the parameters in Eq. (15.9) must becalculated by assuming that the 0 value will represent the loose condition.

However, when piles are installed by rotary drilling under a bentonite slurry for stabilizingthe sides, it may be assumed that the 0 value used to calculate both the skin friction and baseresistance will correspond to the undisturbed soil condition (Tomlinson, 1986).

The assumption of loose conditions for calculating skin friction and base resistance meansthat the ultimate carrying capacity of a bored pile in a cohesionless soil will be considerably lowerthan that of a pile driven in the same soil type. As per De Beer (1965), the base resistance qb of abored and cast-in-situ pile is about one third of that of a driven pile.

624 Chapter 15

We may write,

qb (bored pile) = (1/3) qh (driven pile)

So far as friction load is concerned, the frictional parameter may be calculated by assuming avalue of 0 equal to 28° which represents the loose condition of the soil.

The same Eq. (15.9) may be used to compute Qu based on the modifications explained above.

15.13 MEYERHOF'S METHOD OF DETERMINING Qb FOR PILES IN SAND

Meyerhof (1976) takes into account the critical depth ratio (LJd) for estimating the value of Qb,Fig. 15.10 shows the variation of LJd for both the bearing capacity factors Nc and N as a functionof 0. According to Meyerhof, the bearing capacity factors increase with Lh/d and reach a maximumvalue at LJd equal to about 0.5 (LJd), where Lb is the actual thickness of the bearing stratum. Forexample, in a homogeneous soil (15.6c) Lb is equal to L, the actual embedded length of pile;whereas in Fig. 15.6b, Lh is less than L.

1000

10 20 30

Angle of internal friction in degrees

40 45

Figure 15.10 Bearing capacity factors and critical depth ratios LJd for driven piles(after Meyerhof, 1976)


As per Fig. 15.10, the value of LJd is about 25 for 0 equal to 45° and it decreases with adecrease in the angle of friction 0. Normally, the magnitude of Lbld for piles is greater than0.5 (LJd) so that maximum values ofNc and TV may apply for the calculation of qb, the unit bearingpressure of the pile. Meyerhof prescribes a limiting value for qb, based on his findings on static conepenetration resistance. The expression for the limiting value, qbl, is

for dense sand: qbl = 50 A^ tan 0 kN/m2 (15.19a)

for loose sand: qbl = 25 Nq tan 0 kN/m2 (15.19b)

where 0 is the angle of shearing resistance of the bearing stratum. The limiting qbl values given byEqs (15.9a and b) remain practically independent of the effective overburden pressure andgroundwater conditions beyond the critical depth.

The equation for base resistance in sand may now be expressed as

Qb=«oNqAb*<*uAb (15-20)

where q'o = effective overburden pressure at the tip of the pile LJd and N = bearing capacity factor(Fig. 15.10).

Eq. (15.20) is applicable only for driven piles in sand. For bored cast-in-situ piles the value ofqb is to be reduced by one third to one-half.

Clay Soil (0 = 0)The base resistance Qb for piles in saturated clay soil may be expressed as

Qb=NccuAb=9cuAb (15.21)

where Nc = 9, and cu = undrained shear strength of the soil at the base level of the pile.

15.14 VESIC'S METHOD OF DETERMINING Qb

The unit base resistance of a pile in a (c - 0) soil may be expressed as (Vesic, 1977)

3b = cNc+3'oN*q (15.22)

where c = unit cohesionq'0 = effective vertical pressure at the base level of the pile

N*c and N* = bearing capacity factors related to each other by the equation

A£ = (#;-i)cot0 (15.23)As per Vesic, the base resistance is not governed by the vertical ground pressure q'Q but by the

mean effective normal ground stress crm expressed as

l+2Ko ,am = —-^-3- 3o (15.24)

in which KO = coefficient of earth pressure for the at rest condition = 1 - sin 0.Now the bearing capacity in Eq. (15.22) may be expressed as

=<+^< (15-25)

626 Chapter 15

An equation for AT f fcan be obtained from Eqs. (15.22), (15.24) and (15.25) as

3/V*N° = 1 + 2K (15.26)

o

Vesic has developed an expression for N*a based on the ultimate pressure needed to expand aspherical cavity in an infinite soil mass as

where a, = , a~ = I ^ I tan^ , a-, ~ *""""" r ancj yy - tan2(45° + <b!2)1 3-sin^ 2 \2 ) 3 (1 + sin^) ^

According to Vesic

l + /rA

£,

(15.28)

lr - rigidity index =

where /rr = reduced rigidity index for the soil

A = average volumetric strain in the plastic zone below the pile point

Es = modulus of elasticity of soil

G = shear modulus of soil

j. = Poisson's ratio of soil

Figures 15.1 1 (a) and 15.1 l(b) give plots of A^ versus 0, and N*c versus 0for various values oflrr respectively.

The values of rigidity index can be computed knowing the values of shear modulus G^ and theshear strength s (= c + q'o tan 0).

When an undrained condition exists in the saturated clay soil or the soil is cohesionless and isin a dense state we have A = 0 and in such a case /. = Irr.

For 0 = 0 (undrained condition), we have

A7 = 1.33(ln/,r + l) + !+l (15.30)

The value of lf depends upon the soil state, (a) for sand, loose or dense and (b) for clay low,medium or high plasticity. For preliminary estimates the following values of lr may be used.

Soil type lr

Sand (Dr = 0.5-0.8) 75-150Silt 50-75Clay 150-250


(a)

1000

CQ

100

10

XXX ' /

10 20 30Angle of internal friction 0°

40

500400

200

1006040

2010

50

(b)

1000

100

10

10 20 30Angle of internal friction 0C

40

/,,=

200

1006040

20

10

50

Figure 15.11 (a) Bearing capacity factor^ N*a (b) Bearing capacity factor AT(Veslc, 1977)

628 Chapter 15

Table 15.1 Bearing capacity factors A/* and N* by Janbu

V =0°0

5

10

20

30

35

40

45

75°

N\ N*c

1.00

1.50

2.25

5.29

13.60

23.08

41.37

79.90

5.74

6.25

7.11

11.78

21.82

31.53

48.11

78.90

90°

N\ N*c

1.00

1.57

2.47

6.40

18.40

33.30

64.20134.87

5.74

6.49

8.34

14.83

30.14

46.12

75.31

133.87

105°

N\ N*c

1.00

1.64

2.71

7.74

24.90

48.04

99.61

227.68

5.74

7.33

9.70

18.53

41.39

67.18

117.52226.68

15.15 JANBU'S METHOD OF DETERMINING Qb

The bearing capacity equation of Janbu (1976) is the same as Eq. (15.22) and is expressed as

Qb=Wc+VoN^Ab (15-31)

The shape of the failure surface as assumed by Janbu is similar to that given in Fig. 15.6(b).Janbu's equation for N* is

F i 1= tan 0+ VI + tan2 0 (15.32)

where i//= angle as shown in Fig. 15.6(b). This angle varies from 60° in soft compressible soil to105° in dense sand. The values for N*c used by Janbu are the same as those given by Vesic(Eq. 15.23). Table 15.1 gives the bearing capacity factors of Janbu.

Since Janbu's bearing capacity factor N* depends on the angle y/, there are two uncertaintiesinvolved in this procedure. They are

1. The difficulty in determining the values of i/^for different situations at base level.

2. The settlement required at the base level of the pile for the full development of a plasticzone.

For full base load Qb to develop, at least a settlement of about 10 to 20 percent of the pilediameter is required which is considerable for larger diameter piles.

15.16 COYLE AND CASTELLO'S METHOD OF ESTIMATING Qb IN SAND

Coyle and Castello (1981) made use of the results of 24 full scale pile load tests driven in sand forevaluating the bearing capacity factors. The form of equation used by them is the same as Eq. (15.6)which may be expressed as

Qb=VoNqAi, (15.33)

where q' - effective overburden pressure at the base level of the pileN* = bearing capacity factor

Coyle and Castello collected data from the instrumented piles, and separated from the totalload the base load and friction load. The total force at the top of the pile was applied by means of ajack. The soil at the site was generally fine sand with some percentage of silt. The lowest and the


JO

Bearing capacity factor Nq*

20 40 60 80100 200 (log scale)

Figure 15.12 A/* versus Lid (after Coyle and Castello, 1981

highest relative densities were 40 to 100 percent respectively. The pile diameter was generallyaround 1.5 ft and pile penetration was about 50 ft. Closed end steel pipe was used for the tests insome places and precast square piles or steel H piles were used at other places.

The bearing capacity factor N* was evaluated with respect to depth ratio Lid in Fig. 15.12 forvarious values of </>.

15.17 THE ULTIMATE SKIN RESISTANCE OF A SINGLE PILE INCOHESIONLESS SOIL

Skin Resistance (Straight Shaft)The ultimate skin resistance in a homogeneous soil as per Eq. (15.9) is expressed as

In a layered system of soilthen be expressed as

qo,

(15.34a)

Ks and 8 vary with respect to depth. Equation (15.34a) may

Q= (15.34b)

where q'o, KS and <5 refer to thickness dz of each layer and P is the perimeter of the pile.As explained in Section 15.10 the effective overburden pressure does not increase linearly

with depth and reaches a constant value beyond a particular depth Lc, called the critical depth whichis a function of 0. It is therefore natural to expect the skin resistance fs also to remain constantbeyond depth L . The magnitude of L may be taken as equal to 2Qd.

630 Chapter 1 5

Table 15.2 Values of Ks and 8 (Broms, 1966)

Pile material

Values of Ks

Low Dr High Dr

Steel

Concrete

Wood

20°

3/40

2/30

0.5

1.0

1.5

1.0

2.0

4.0

3.0

2.5

- 2.0

1.5

l.Q

_l 1 1 1

-

-

^/

1 1 1 1

Drivenpiles /

/i i i i

i i i i_

/:

/ '--

i i i r

1.6

1.2

0.8

0.4

i i- D

-

-

i i

i i i i

riven pil\

\

1s

1 1 1 1

1 1 1 1

e

V '

i i

-

/ Jacked.

x^- Boredpiles

i i i i i i28 33 38 43

0°(a)

30 35 400°(b)

25'

0.5 1.0 1.5 2.0Pile taper angle, a>°

(c)

Figure 15.13 Values of Ks tan 8 in sand as per (a) Poulos and Davis 1980, (b)Meyerhof, 1 976 and (c) taper factor Fw (after Nordlund, 1 963)

Eq. (15.17) can be used for determining the_critical length Lc for any given set of values ofand d. (Xcan be calculated from Eq. (15.34) if A^ and 8 are known.


The values of Ks and 5 vary not only with the relative density and pile material but also withthe method of installation of the pile.

Broms ( 1 966) has related the values of Ks and 8 to the effective angle of internal friction 0 ofcohesionless soils for various pile materials and relative densities (Dr) as shown in Table 15.2. Thevalues are applicable to driven piles. As per the present state of knowledge, the maximum skinfriction is limited to 1 10 kN/m2 (Tomlinson, 1986).

Eq. (15.34) may also be written as

Q = Pq'oj3dz (15.35)o

where, /3 = Ks tan 8 .Poulos and Davis, (1980) have given a curve giving the relationship between /3and 0° which

is applicable for driven piles and all types of material surfaces. According to them there is notsufficient evidence to show that /? would vary with the pile material. The relationship between ft and0 is given in Fig. 15.13(a). For bored piles, Poulos et al, recommend the relationship given byMeyerhof (1976) between 0 and 0 (Fig. 15.13(b)).

Skin Resistance on Tapered PilesNordlund (1963) has shown that even a small taper of 1° on the shaft gives a four fold increase inunit friction in medium dense sand under compression loading. Based on Nordlund's analysis,curves have been developed (Poulos and Davis, 1980) giving a relationship between taper angle ofand a taper correction factor Fw, which can be used in Eq. (15.35) as

Qf= FvPq'oPte (15.36)o

Eq. (15.36) gives the ultimate skin load for tapered piles. The correction factor Fw can beobtained from Fig. 15.13(c). The value of 0 to be used for obtaining Fw is as per Eq. (15.18a) fordriven piles.

15.18 SKIN RESISTANCE Qf BY COYLE AND CASTELLO METHOD (1981)

For evaluating frictional resistance, Q~ for piles in sand, Coyle and Castello (1981) made use of theresults obtained from 24 field tests on piles. The expression for <2,is_the one given in Eq. (15.34a).They developed a chart (Fig. 15.14) giving relationships between Ks and 0 for various Lid ratios.The angle of wall friction 8 is assumed equal to 0.80. The expression for Qfis

Qf=Asq'oKstanS (15.34a)

where q'o = average effective overburden pressure and S = angle of wall friction = 0.80.The value of Ks can be obtained Fig. 15.14.

15.19 STATIC BEARING CAPACITY OF PILES IN CLAY SOILEquation for Ultimate Bearing CapacityThe static ultimate bearing capacity of piles in clay as per Eq. (15.12) is

Qu = Qb +Qf = cbNcAb+acuAs (15.37)

632 Chapter 1 5

Earth pressure coefficient Ks

j>= 30 32 34 36°

Figure 15.14 Coefficient K versus L/d, 8 = 0.80 (after Coyle and Castello, 1981)

For layered clay soils where the cohesive strength varies along the shaft, Eq. (15.37) may bewritten as

(15.38)

Bearing Capacity Factor Nc

The value of the bearing capacity factor Nc that is generally accepted is 9 which is the valueproposed by Skempton (1951) for circular foundations for a LIB ratio greater than 4. The basecapacity of a pile in clay soil may now be expressed as

Q, = 9c,A, (15 39)*--b b b \L~J.~JSJ

The value of cb may be obtained either from laboratory tests on undisturbed samples or fromthe relationships established between cu and field penetration tests. Eq. (15.39) is applicable for alltypes of pile installations,

Skin Resistance by a-MethodTomlinson (1986) has given some empirical correlations for evaluating a in Eq. (15.37) fordifferent types of soil conditions and Lid ratios. His procedure requires a great deal of judgment ofthe soil conditions in the field and may lead to different interpretations by different geotechnicalengineers. A simplified approach for such problems would be needed. Dennis and Olson (1983b)


made use of the information provided by Tomlinson and developed a single curve giving therelationship between a and the undrained shear strength cu of clay as shown in Fig. 15.15.

This curve can be used to estimate the values of a for piles with penetration lengths less than30 m. As the length of the embedment increases beyond 30 m, the value of a decreases. Piles ofsuch great length experience elastic shortening that results in small shear strain or slip at great depthas compared to that at shallow depth. Investigation indicates that for embedment greater than about50 m the value of a from Fig 15.15 should be multiplied by a factor 0.56. For embedments between30 and 50 m, the reduction factor may be considered to vary linearly from 1.0 to 0.56 (Dennis andOlson, 1983a, b)

Skin Resistance by A-MethodVijayvergiya and Focht (1972) have suggested a different approach for computing skin load QAovsteel-pipe piles on the basis of examination of load test results on such piles. The equation is of theform

Qf = A(q'0 + 2cM) As (15.40)

where A = frictional capacity coefficient,q'o = mean effective vertical stress between the ground surface and pile tip.

The other terms are already defined. A is plotted against pile penetration as shown inFig. 15.16.

Eq. (15.40) has been found very useful for the design of heavily loaded pipe piles for offshorestructures.

/MVIethod or the Effective Stress Method of Computing Skin Resistance

In this method, the unit skin friction fs is defined as

fs = Ks tanSq'o = (3~q'o (15.41)

where ft = the skin factor = Ks tan <5, (15.42a)

K = lateral earth pressure coefficient,

Undrained cohesion cu, kN/m2

Figure 15.15 Adhesion factor a. for piles with penetration lengths less than 50 m in clay.(Data from Dennis and Olson 1983 a, b; Stas and Kulhawy, 1984)

634 Chapter 1 5

Value of AI.2 0.3 0.4 0.5

Figure 15.16 Frictional capacity coefficient A vs pile penetration(Vijayvergiya and Focht, 1972)

8 = angle of wall friction,q'o = average effective overburden pressure.

Burland (1973) discusses the values to be used for )3 and demonstrates that a lower limit forthis factor for normally consolidated clay can be written as

~ (15.42b)

(15.42c)

(15.42d)

As per Jaky (1944)

therefore /3 = (1 - sin 0') tan 0'where 0' = effective angle of internal friction.

Since the concept of this method is based on effective stresses, the cohesion intercept on aMohr circle is equal to zero. For driven piles in stiff overconsolidated clay, Ks is roughly 1.5 timesgreater than KQ. For overconsolidated clays KO may be found from the expression

Ko = (1 - sin <!)'}TJR~ (15.42e)

where R - overconsolidation ratio of clay.For clays, 0' may be taken in the range of 20 to 30 degrees. In such a case the value of |3 in

Eq. (15.42d) varies between 0.24 and 0.29.


Meyerhof 's Method (1976)

Meyerhof has suggested a semi-empirical relationship for estimating skin friction in clays.For driven piles:

fs = \.5cu tan 0' (15.43)

For bored piles:

fs=cutanp (15.44)

By utilizing a value of 20° for </>' for the stiff to very stiff clays, the expressions reduce toFor driven piles:

/, = 0-55c, (15.45)

For bored piles:

fs=0.36cu (15.46)

In practice the maximum value of unit friction for bored piles is restricted to 100 kPa.

15.20 BEARING CAPACITY OF PILES IN GRANULAR SOILSBASED ON SPT VALUEMeyerhof (1956) suggests the following equations for single piles in granular soils based on SPTvalues.

For displacement piles:

Qu = Qb + Qf = 40Ncor(L/d)Ab + 2NcorAs (15.47a)

for //-piles:

Qu = 40Ncor(L/d)Ab + NcorAs (15.47b)

where qb = 40 Ncor(Lld) < 400 Ncor

For bored piles:

Qu=MNcorAb+°*™corA, (15-48)

where Qu = ultimate total load in kNNcor = average corrected SPT value below pile tipNcor = corrected average SPT value along the pile shaft

Ab = base area of pile in m2 (for H-piles including the soil between the flanges)As = shaft surface area in m2

In English units Qu for a displacement pile is

0u(kip) =Qb+Qf= O.WNcor(L/d)Ab +O.Q4NcorAs (15.49a)

where Ab = base area in ft2 and As = surface area in ft2

and 0.80 , AAA < SNcorAb(kip) (15.49b)a

636 Chapter 15

A minimum factor of safety of 4 is recommended. The allowable load Qa is

Q ==±U* 4

(15.50)

Example 15.1

A concrete pile of 45 cm diameter was driven into sand of loose to medium density to a depth of15m. The following properties are known:

(a) Average unit weight of soil along the length of the pile, y = 17.5 kN/m3 , average 0 = 30°,

(b) average Ks = 1.0 and 8= 0.750.

Calculate (a) the ultimate bearing capacity of the pile, and (b) the allowable load withFs = 2.5. Assume the water table is at great depth. Use Berezantsev's method.

Solution

From Eq. (15.9)

Qu=Qb+Qf= q'0AbNq + q'0AKs tan S

where ' = L = 17.5 x 15 = 262.5 kN/m2

= 131.25 kN/m2J O f\ '

A, = —~ x0.452= 0.159 m2b 4

Qu

15mQf

Sandy = 17.5 kN/m3

0 = 30°£,= 1.0

Qb

Figure Ex. 15.1

ars

Highlight


As =3.14x0.45x15 = 21.195 m2

S = 0.750 = 0.75x30 = 22.5°

tan 8 =0.4142

L 15From Fig. 15.9, TV for — = — — = 33.3 and 0 = 30° is equal to 16.5.q a 0.45Substituting the known values, we have

Qu =Qb+Qf =262.5x0.159x16.5 + 131.25x21.195x1.0x0.4142

= 689+ 1152 = 1841 kN

Example 15.2Assume in Ex. 15.1 that the water table is at the ground surface and vt = 18.5 kN/m3. All the other

*" * Sal

data remain the same. Calculate Qu and Qa.

Solution

Water table at the ground surface ysat =18.5 kN/m3

rb=rM~rw=l 8-5 - 9.8 1 = 8.69 kN/m3

^=8.69x15 = 130.35 kN/m2

q'Q = -x 130.35 = 65.18 kN/m2

Substituting the known values

Qu = 1 30.35 x 0.159 x 16.5 + 65.18 x 21.195 x 1.0 x 0.4142

= 342 + 572 = 914 kN

914e a= — = 366 kN

Note: It may be noted here that the presence of a water table at the ground surface incohesionless soil reduces the ultimate load capacity of pile by about 50 percent.

Example 15.3A concrete pile of 45 cm diameter is driven to a depth of 16 m through a layered system of sandysoil (c = 0). The following data are available.

Top layer 1: Thickness = 8 m, yd = 16.5 kN/m3, e = 0.60 and 0 = 30°.Layer 2: Thickness = 6 m, yd = 15.5 kN/m3, e - 0.65 and 0 = 35°.Layer 3: Extends to a great depth, yd = 16.00 kN/m3, e = 0.65 and 0 = 38°.

638 Chapter 15

Q

T

A L , =

AL2 =

i

AL3 =

Layer 1Sand

= 8m

Layer 2Sand

= 6 m

= 2 m Laygr 3

Sand

T

yd] = 16.5 kN/m3

0 = 30°, e = 0.6

yd2= 15. 5 kN/m3

0 = 35°, e = 0.65

yd3= 16.0 kN/m3

0 = 38°, e = 0.65

Figure Ex. 15.3

Assume that the value of <5in all the layers of sand is equal to 0.750. The value of KS for each

layer as equal to half of the passive earth pressure coefficient. The water table is at ground level.Calculate the values of Qu and Qa with Fs = 2.5 by the conventional method for Qf and

Berezantsev's method for Qb.

Solution

The soil is submerged throughout the soil profile. The specific gravity G^ is required for calculating

'sat'

Y G(a) Using the equation Yd = ~—L' calculate G^ for each layer since yd, Yw and e are known.

Y (G(b) Using the equation 7sat = : , calculate ysat for each layer and then yb = ysat - yw for

each layer.(c) For a layered system of soil, the ultimate load can be determined by making use of

Eq. (15.9). Now

Qu=Qb+Qf= q',NqAb + P L ^ Ks tan^ALo

(d) q'Q at the tip of the pile is


(e) q'Q at the middle of each layer is

4oi=-AL,rM

-

(f) A/j = 95 for 0 = 38° and — = — = 33.33 from Fig. 15.9.

(g) Afc = 0.159 m2 ,P= 1.413m.

- 1 / ^ 1 -(h) Ks=-tai\2\45°+—\=—Kp. Ks for each layer can be calculated.

(i) 8 = 0.75</>. The values of tan 8 can be calculated for each layer.The computed values for all the layers are given below in a tabular form.

Layer no. Gs

1.2.

3.

2.69

2.61

2.69

From middle of layer 3

At the tip of pile

Yt,kN/m 3

10.36

9.57

10.05

to tip of pile

4'o =

kN/m2

41.44 1.5

111.59 1.845

150.35 2.10

= 10.05

160.40 kN/m2

tan 8 A/,

m

0.414 8

0.493 6

0.543 2

<2W =(160.4x95x0.159 + 1.413(41.44x1.5x0.414x8

+ 1 1 1.59 x 1.845 x 0.493 x 6 + 150.35 x 2.10 x 0.543 x 2)

= 2423 + 1636 = 4059 kN

4059Q =a 2.5 2.5

Example 15.4If the pile in Ex. 15.2 is a bored and cast-in-situ, compute Qu and Qa. All the other data remain thesame. Water table is close to the ground surface.

Solution

Per Tomlinson (1986), the ultimate bearing capacity of a bored and cast-in-situ-pile in cohesionlesssoil is reduced considerably due to disturbance of the soil. Per Section 15.12, calculate the baseresistance for a driven pile and take one-third of this as the ultimate base resistance for a bored andcast-in-situ pile.

For computing 8, take 0 = 28° and K = 1.0 from Table 15.2 for a concrete pile.

640 Chapter 15

Base resistance for driven pile

For 0 = 30°, Nq = 16.5 from Fig. 15.9.

Ah = 0.159m2

q'Q = 130.35 kN/m2 (From Ex. 15.2)

Qb = 130.35 x 0.159 x 16.5 = 342 kN

For bored pile

e,=-

Skin load

Qf =A

For 0= 28°, 5= 0.75 x 28 = 21°, tan<5 = 0.384

As = 21.195 m2 (from Ex. 15.2)

^=65.18 kN/m2 (Ex. 15.2)

Substituting the known values,

Qf = 21.195 x 65.18 x 1.0 x 0.384 = 530 kN

Therefore, Qu = 114 + 530 = 644 kN

644

Example 15.5Solve the problem given in Example 15.1 by Meyerhof 's method. All the other data remain thesame.

Solution

Per Table 9.3, the sand in-situ may be considered in a loose state for 0 = 30°. The corrected SPTvalue NCQT =10.

Point bearing capacityFrom Eq. (15.20)

From Eq. (15.19 b)

qbl = 25 N tan 0 kN/m2

Now From Fig. 15. 10 Nq = 60 for 0 = 30°

q'o = y L = 17.5 x 15 = 262.5 kN/m2

qb = 262.5 x 60 = 15,750 kN/m2

a, , = 25 x 60 x tan 30° = 866 kN/m2

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Hence the limiting value for qb = 866 kN/m2

314Now,Qb =Abqb= Abqb = — x(0.45)2x 866 = 138 kN

Frictional resistancePer Section 15.17, the unit skin resistance fs is assumed to increase from 0 at ground level to alimiting value of/5/ at Lc =2Qd where Lc = critical depth and d = diameter. Therefore Lc = 20 x 0.45= 9m

Now/s, = q'0Ks tan 8=yLcKs tan8

Given: y = 17.5 kN/m3, Lc = 9 m, Ks = 1.0 and 8= 22.5° m.

Substituting and simplifying we have

fsl = 17.5 x 9 x 1.0 x tan 22.5 = 65 kN/m2

The skin load Qf = Qfl + Qf2 = ^fslPLc+Pfsl(L-Lc)

Substituting Qf = - x 65 x 3.14 x 0.45 x 9 + 3.14 x 0.45 x 65(15 - 9)

= 413 + 551 = 964 kN

The failure load Qu is

Qu = Qb + Qf= 138+ 964 = U02 kN

with Fc = 2.5,

= 440 kN~a 2.5

Example 15.6Determine the base load of the problem in Example 15.5 by Vesic's method. Assume lr = lrr = 50.Determine Qa for Fs = 2.5 using the value of QAn Ex 15.5.

Solution

From Eq. (15.25) for c = 0 we have

From Eq. (15.24)

l + 2KQ , _ 1 + 2(1-shift) ,m 3 3

^=15x17.5 = 262.5 kN/m2

1 + 2(1-sin 30°)O". = x 262.5 = 175 kN/m2

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642 Chapter 15

From Fig. 15.1 la, N*a = 36 for 0 = 30 and lr = 50Substituting

qb = 175 x 36 = 6300 kN/m2

3.14Qb = Abqb = ~— x (0.45)2 x 6300 = 1001 kN

Qu = Qb + Qf = 1001 + 964 = 1,965 kN

2.5

Example 15.7Determine the base load of the problem in Example 15.1 by Janbu's method. Use if/ = 90Determine Qa for FS = 2.5 using the Q, estimated in Example 15.5.

Solution

From Eq (15.31), for c = 0 we have

For 0 = 30° and y = 90°, we have AT * = 18.4 from Table 15.1. q'Q = 262.5 kN/m2 as inEx. 15.5.

Therefore qb = 262.5 x 18.4 = 4830 kN/m2

Qb = Abqb = 0.159 x 4830 = 768 kN

Qu = Qb+Qf= 768 + 964 = 1732 kN

2.5

Example 15.8Estimate Qb, Q* Qu and <2aby the Coyle and Castello method using the data given in Example 15.1.

Solution

Base load Qb from Eq. (15.33)

Vb = 4oN*q

From Fig. 15.12, N*q = 29 for 0 = 30° and Lid = 33.3

q'o = 262.5 kN/m2 as in Ex. 15.5

Therefore qb = 262.5 x 29 = 7612 kN/m2

Qb =Ab^b =0.159x7612-1210 kN

From Eq. (15.34a)

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where As = 3.14 x 0.45 x 15 = 21.2 m2

q' = - x 262.5 = 13 1.25 kN/m2v0 2

S = 0.80 = 0.8x30° =24°

From Fig. 15.14, Ks = 0.35 for </> = 30° and Lid = 33.3

Therefore Qf = 2 1.2x13 1.25x0.35 tan 24° =434 kN

Qu=Qb+Qf= 1210 + 434 = 1644 kN

6442.5

=658 kN

Example 15.9Determine Qb, Qf Qu and Qa by using the SPT value for 0 = 30° from Fig. 12.8.

Solution

From Fig. 12.8, Ncor = 10 for 0 = 30°. Use Eq. (15.47a) for Qu

Qu=Qb+Qf=WNcor± Ab+2NcorAs

where (2 f t<Gw = 400 rA6

Given: L = 15 m, d = 0.45 m, Ab = 0.159 m2, As = 21.2 m2

Q, =40x10 x — x 0.159 = 2120 kN* 0.45

Q w = 400x10x0.1 59 = 636 kN

Since 0^, > <2W, use <2W

Qf =2x10x21.2 = 424 kN

Now <2M = 636 + 424 = 1060 kN

Q =1^ = 424 kNa 2.5

Example 15.10Compare the values of Qb, (Land Qa obtained by the different methods in Examples 15.1, and 15.5through 15.9 and make appropriate comments.

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644 Chapter 15

Comparison

The values obtained by different methods are tabulated below.

MethodNo

1.2.

3.

4.

5.

6.

ExampleNo

15.1

15.5

15.6

15.7

15.8

15.9

Investigator

Berezantsev

Meyerhof

Vesic

Janbu

Coyle & Castello

Meyerhof(Based on SPT)

abkN

689

138

1001

768

1210

636

OfkN

1152

964

964

964

434

424

QUkN

1841

1102

1965

1732

1644

1060

Qa(Fs = 2.5)

kN

736

440

786

693

658

424

CommentsIt may be seen from the table above that there are wide variations in the values of Qb and Q,betweenthe different methods.

Method 1 Tomlinson (1986) recommends Berezantsev's method for computing Qb as thismethod conforms to the practical criteria of pile failure. Tomlinson does notrecommend the critical depth concept.

Method 2 Meyerhof's method takes into account the critical depth concept. Eq. (15.19) isbased on this concept. The equation [Eq. (15.20)] qb - q'QNq does not consider thecritical depth concept where q' = effective overburden pressure at the pile tip levelof the pile. The value of Qb per this equation is

Qb = qbAb= 15,750 x 0.159 = 2504 kN

which is very high and this is close to the value of Qb (= 2120 kN) by the SPTmethod. However Eq. (15.19 b) gives a limiting value for Qb=l38 kN (here the sandis considered loose for 0 = 30.QAS computed by assuming (Xincreases linearly with depth from 0 at L = 0 to Qfl atdepth Lc = 20d and then remains constant to the end of the pile.Though some investigators have accepted the critical depth concept for computingQb and (X, it is difficult to generalize this concept as applicable to all types ofconditions prevailing in the field.

Method 3 Vesic's method is based on many assumptions for determining the values of /., Irr,<7m, N*a etc. There are many assumptions in this method. Are these assumptions validfor the field conditions? Designers have to answer this question.

Method 4 The uncertainties involved in Janbu's method are given in Section 15.15 and as suchdifficult to assess the validity of this method.

Method 5 Coyle and Castello's method is based on full scale field tests on a number of drivenpiles. Their bearing capacity factors vary with depth. Of the first five methods listedabove, the value of Qb obtained by them is much higher than the other four methodswhereas the value of QAS very much lower. But on the whole the value of Qu is lowerthan the other methods.

Method 6 This method was developed by Meyerhof based on SPT values. The Qb value(=2120kN) by this method is very much higher than the preceding methods,


whereas the value of Q,is the lowest of all the methods. In the table given above thelimiting value of Qbl = 636 kN is considered. In all the six methods Fs = 2.5 has beentaken to evaluate Qa whereas in method 6, Fs = 4 is recommended by Meyerhof.

Which Method to UseThere are wide variations in the values of Qb, <2« Qu and Qa between the different methods. Therelative proportions of loads carried by skin friction and base resistance also vary between themethods. The order of preference of the methods may be listed as follows;

Preference No.

12

3

4

5

6

Method No.

1

2

5

63

4

Name of the investigator

Berezantsev for Qh

Meyerhof

Coyle and Castello

Meyerhof (SPT)

Vesic

Janbu

Example 15. 11

A concrete pile 18 in. in diameter and 50 ft long is driven into a homogeneous mass of clay soil ofmedium consistency. The water table is at the ground surface. The unit cohesion of the soil underundrained condition is 1050 lb/ft2 and the adhesion factor a = 0.75. Compute Qu and Q withF, = 2.5.

SolutionGiven: L = 50 ft, d = 1.5 ft, cu = 1050 lb/ft2, a = 0.75.

From Eq. (15.37), we have

Q = Qh+Qf =chN A.+A ac*~-u *--b *-•/ b c b s u

where, cb = cu = 1050 lb/ft2; Nc =%Ab= 1.766 ft2; As = 235.5 ft2

Substituting the known values, we have

1050x9x1.766 235.5x0.75x1050_" ~ 1000 1000

= 16.69 + 185.46 = 202.15 kips

Q =a 2.5

Example 15.12A concrete pile of 45 cm diameter is driven through a system of layered cohesive soils. The lengthof the pile is 16m. The following data are available. The water table is close to the ground surface.

Top layer 1: Soft clay, thickness = 8 m, unit cohesion cu = 30 kN/m2 and adhesionfactor a = 0.90.

Layer 2: Medium stiff, thickness = 6 m, unit cohesion cu = 50 kN/m2 and a = 0.75.

646 Chapter 1 5

Layer 3: Stiff stratum extends to a great depth, unit cohesion cu = 105 kN/m2 and

a = 0.50.

Compute Qu and Qa with Fs = 2.5.

Solution

Here, the pile is driven through clay soils of different consistencies.The equations for Qu expressed as (Eq. 15.38) yield

Here, cb = cu of layer 3, P = 1.413 m,Ab = 0.159 m2


Qu =9x105x0.159 + 1.413(0.90x30x8

+ 0.75x50x6 + 0.50x105x2)

= 150.25 + 77 1.5 = 92 1.75 kN

= =5

2.5

Q

Tj

AL, =

1

AL2 =

1

AL3:

Layer 1

= 8m

Layer 2

- 6 m

: 2 m Layer 3

T

Soft clay

cu = 30 kN/m2

a = 0.90

— 45 cm

Medium stiff clay

cu = 50 kN/m2

a = 0.75

Stiff clay cu= 105 kN/m2

' a = 0.50

Figure Ex. 15.12


Example 15.13A precast concrete pile of size 18 x 18 in is driven into stiff clay. The unconfmed compressivestrength of the clay is 4.2 kips/ft2. Determine the length of pile required to carry a safe working loadof 90 kips with Fs = 2.5.

Solution

The equation for Qu is

we have

Qu = 2.5 x 90 = 225 kips,

Nc = 9, cu = 2.1 kips/ft2

a = 0.48 from Fig. 15.15, cu = cu = 2.1 kips/ft2, Ab = 2.25 ft2

Assume the length of pile = L ft

Now,A5 = 4 x 1.5L = 6L


225 = 9 x 2.1 x 2.25 + 0.48 x 2.1 x 6L

or 225 = 42.525 + 6.05L

Simplifying, we have

225-42.525L =

6.05= 30.2 ft

/w\

. 118 x 18 in.

. Stiff clayf cu = 2.1 kips/ft2

1 a = 0.48

Figure Ex. 15.13

648 Chapter 15

Example 15.14

For the problem given in Example 15.11 determine the skin friction load by the A-method. All theother data remain the same. Assume the average unit weight of the soil is 1 10 lb/ft3. Use Qb given inEx. 15.11 and determine Qa for FS = 2.5.

SolutionPerEq. (15.40)

Q = A(q" +2c )Ax-'f ^" O U ' S

gj=-x50x 110 = 2,750 lb/ft2

Depth = 50 ft = 15.24m

From Fig. 15.16, A = 0.2 for depth L = 15.24 m

0.2(2750 + 2x 1050) x 235.5 „ „ „ , , , .Now Or = — - - - - = 228.44 kipsf 1000

Now Qu = Qb + Qf = 1 6.69 + 228.44 - 245 kips

Qa=- = 98 kips

Example 15.15A reinforced concrete pile of size 30 x 30 cm and 10m long is driven into coarse sand extending toa great depth. The average total unit weight of the soil is 18 kN/m3 and the average Ncor value is 15.Determine the allowable load on the pile by the static formula. Use F^ = 2.5. The water table is closeto the ground surface.

SolutionIn this example only the /V-value is given. The corresponding </> value can be found from Fig. 12.8which is equal to 32°.

Now from Fig. 15.9, for <j> = 32°, and — = —- = 33.33, the value of Na = 25.a 0.3 q

Ab =0.3x0.3 = 0.09m2,

Av =10x4x0.3= 12m2

£=0.75x32 = 24°, tan 8 = 0.445

The relative density is loose to medium dense. From Table 15.2, we may take

^ = l + -(2-l) = 1.33

Now, Qu = q'QNqAb + q'QKs \.mSAs

yb = rsat -yw = 18.0-9.81 - 8.19 kN/m3

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Qu

//A\\

J

1

fi*

0.30 x 0.30 m

Medium dense sandy sa t=18kN/m3

^ = 32°

Figure Ex. 15.15

q'Q = ybl = 8.19 x 10 = 81.9 kN/m2

2 2


Qu = 81.9 x 25 x 0.09 + 40.95 x 1.33 x 0.445 x 12

= 184 + 291 = 475 kN

475

Example 15.16Determine the allowable load on the pile given in Ex. 15.15 by making use of the SPT approach byMeyerhof.

Solution

Per Ex. 15.15, Ncor= 15

Expression for Qu is (Eq. 15. 47 a)

Qu = Qb + Qf = 4QNcor(L/d)Ab+2NcorAs

Here, we have to assume Ncor = Ncor =15

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650 Chapter 15

Ab = 0.3x 0.3 = 0.09 m2 , As = 4x0.3x10= 12 m2

Substituting, we have

G, =40x15 — x 0.09 = 1800 kN* 0.3

Qbl = 400Ncor x Ab = 400 x 15 x 0.09 = 540 kN < Qb

Hence, Qbl governs.

Qf = 2NcorAs = 2 x 15 x 12 = 360 kN

A minimum Fy = 4 is recommended, thus,

,gLtg/.= 540 + 360^"fl A A

Example 15.17Precast concrete piles 16 in. in diameter are required to be driven for a building foundation. Thedesign load on a single pile is 100 kips. Determine the length of the pile if the soil is loose tomedium dense sand with an average Ncor value of 15 along the pile and 21 at the tip of the pile. Thewater table may be taken at the ground level. The average saturated unit weight of soil is equal to120 lb/ft3. Use the static formula and Fs = 2.5.

Solution

It is required to determine the length of a pile to carry an ultimate load ofQu = 2.5x 100 = 250 kips.

The equation for Qu is

The average value of 0 along the pile and the value at the tip may be determined fromFig. 12.8.

For Ncor = 15, 0 = 32°; for N£or = 21, 0 = 33.5°.Since the soil is submerged

yb =120 -62.4 -57.6 lb/ft3

Now

lb/ft2

314

= 28.8L lb/ft2

x(1.33)2 = 1.39ft2

A =3.14xl.33xL = 4.176Lft2

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16 in/

Medium dense sandNcor=\5ysat=18.51b/ft3

Figure Ex. 15.17

From Fig. 15.9, N = 40 for Lid = 20 (assumed)

From Table 15.2, Kf = 1.33 for the lower side of medium dense sand

S = -x 33.5 = 25.1°, tan S= 0.4694

Now by substituting the known values, we have

(57.6L)x 40x1.39 (28.8L) x 0.469 x 1.33 x (4.1 76L)

1000 1000

= 3.203L + 0.075L2

or Z2 + 42.7 1L- 3333 = 0

Solving this equation gives a value of L = 40.2 ft or say 41 ft.

Example 15.18Refer to the problem in Example 15.17. Determine directly the ultimate and the allowable loadsusing Ncor. All the other data remain the same.

Solution

UseEq. (15.49a)

Qu = Qb + Qf = O.

Given: Ncor = 21, Ncor = 15, d = 16 in, L = 41 ft

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652 Chapter 15

3.14 16_ =139ft2 A =314x lH x 41-171.65 ft2

b 4 12 s 12

Substituting

Q, = 0.80 x 21 —- x 1.39 = 720 < 8 W A. kips*^b 1 TO cor b "

Qbl =8x21x1.39 = 234 kips

The limiting value of Qh = 234 kips

Qf = Q.Q4NajrAs = 0.04 x 15 x 171.65 = 103 kips

Qu = Qb + Qf = 234 + 103 = 337 kips

337

15.21 BEARING CAPACITY OF PILES BASED ON STATIC CONEPENETRATION TESTS (CPT)

Methods of Determining Pile CapacityThe cone penetration test may be considered as a small scale pile load test. As such the results ofthis test yield the necessary parameters for the design of piles subjected to vertical load. The typesof static cone penetrometers and the methods of conducting the tests have been discussed in detailin Chapter 9. Various methods for using CPT results to predict vertical pile capacity have beenproposed. The following methods will be discussed:

1. Vander Veen's method.

2. Schmertmann's method.

Vander Veen's Method for Piles in Cohesionless Soils

In the Vander Veen et al., (1957) method, the ultimate end-bearing resistance of a pile is taken,equal to the point resistance of the cone. To allow for the variation of cone resistance whichnormally occurs, the method considers average cone resistance over a depth equal to three times thediameter of the pile above the pile point level and one pile diameter below point level as shown inFig. 15.17(a). Experience has shown that if a safety factor of 2.5 is applied to the ultimate endresistance as determined from cone resistance, the pile is unlikely to settle more than 15 mm underthe working load (Tomlinson, 1986). The equations for ultimate bearing capacity and allowableload may be written as,

pile base resistance, qb = q (cone) (15.51a)

ultimate base capacity, Qb = Abq (15.51b)

Abqallowable base load, Q — (15.51c)

Fs

where, q = average cone resistance over a depth 4d as shown in Fig. 15.17(a) and Fs = factor ofsafety.


The skin friction on the pile shaft in cohesionless soils is obtained from the relationshipsestablished by Meyerhof (1956) as follows.

For displacement piles, the ultimate skin friction, fs, is given by

fs=^(kPa) (15.52a)

and for H-section piles, the ultimate limiting skin friction is given by

/,=-^-(kPa) (15.52b)

where qc = average cone resistance in kg/cm2 over the length of the pile shaft under consideration.Meyerhof states that for straight sided displacement piles, the ultimate unit skin friction, fs,

has a maximum value of 107 kPa and for H-sections, a maximum of 54 kPa (calculated on all facesof flanges and web). The ultimate skin load is

Qf=Asfs (15.53a)

The ultimate load capacity of a pile is

Qu=Qb+Qf (15.53b)

The allowable load is

(15.53c)

If the working load, Qa, obtained for a particular position of pile in Fig. 15.17(a), is less thanthat required for the structural designer's loading conditions, then the pile must be taken to a greaterdepth to increase the skin friction fs or the base resistance qb.

Schmertmann's Method for Cohesionless and Cohesive Soils

Schmertmann (1978) recommends one procedure for all types of soil for computing the pointbearing capacity of piles. However, for computing side friction, Schmertmann gives two differentapproaches, one for sand and one for clay soils.

Point Bearing Capacity Qb in All Types of Soil

The method suggested by Schmertmann (1978) is similar to the procedures developed byDe Ruiter and Beringen (1979) for sand. The principle of this method is based on the one suggestedby Vander Veen (1957) and explained earlier. The procedure used in this case involves determininga representative cone point penetration value, q , within a depth between 0.7 to 4d below the tiplevel of the pile and Sd above the tip level as shown in Fig. 15. 17(b) and (c). The value of q may beexpressed as

(q cl+q c2)/2 + q 3- £2. (15 54)

where qcl = average cone resistance below the tip of the pile over a depth which may varybetween O.ld and 4d, where d = diameter of pile,

qc2 = minimum cone resistance recorded below the pile tip over the same depth O.ld to 4d,

654 Chapter 15

<7c3 = average of the envelope of minimum cone resistance recorded above the pile tip to aheight of Sd.

Now, the unit point resistance of the pile, qb, is

qb (pile) = qp (cone) (15.55a)

The ultimate base resistance, Qh, of a pile is

Qb = \qp (15.55b)

The allowable base load, O is*~-a

(15.55c)

Method of Computing the Average Cone Point Resistance q

The method of computing qcl, qc2 and qc3 with respect to a typical gc-plot shown in Fig. 15.17(b)and (c) is explained below.

Case 1: When the cone point resistance qc below the tip of a pile is lower than that at the tip(Fig. 15.17(b)) within depth 4d.

=

where qo, qb etc., refer to the points o, b etc, on the gc-profile, qc2 = qc = minimum value below tipwithin a depth of 4d at point c on the ^-profile.

The envelope of minimum cone resistance above the pile tip is as shown by the arrow markalong (15.17b) aefghk.

+ / 2 + d + d + / 2 + d

where qa = qe, qf=qg, qh = qk.

Case 2: When the cone resistance qc below the pile tip is greater than that at the tip within a depth4d. (Fig. 15.1 7(c)).

In this case q is found within a total depth of 0.1 d as shown in Fig. 15.17(c).

qc2 = qo = minimum value at the pile tip itself, qc3 = average of the minimum values along theenvelope ocde as before.

In determining the average qc above, the minimum values qc2 selected under Case 1 or 2 areto be disregarded.

Effect of Overconsolidation Ratio in Sand

Reduction factors have been developed that should be applied to the theoretical end bearing of apile as determined from the CPT if the bearing layer consists of overconsolidated sand. Theproblem in many cases will be to make a reasonable estimate of the Overconsolidation ratio in sand.In sands with a high qc, some conservation in this respect is desirable, in particular for shallowfoundations. The influence of Overconsolidation on pile end bearing is one of the reasons forapplying a limiting value to pile end bearing, irrespective of the cone resistances recorded in the


J,

qp (average)

4d

(a) Vander Veen's method

Limits for Eq. (15.56)

(b) Resistance below pile tip lower than (c) Resistance below pile tip greater thanthat at pile tip within depth 4d that at pile tip within 0.75 depth

Figure 15.17 Pile capacity by use of CPT values (a) Vander Veen's method, and(b,c) Schmertmann's method

bearing layer. A limit pile end bearing of 15 MN/m2 is generally accepted (De Ruiter and Beringen,1979), although in dense sands cone resistance may be greater than 50 MN/m2. It is unlikely that indense normally consolidated sand ultimate end bearing values higher than 15 MN/m2 can occur butthis has not been adequately confirmed by load tests.

Design CPT Values for Sand and ClayThe application of CPT in evaluating the design values for skin friction and bearing asrecommended by De Ruiter and Beringen (1979) is summarized in Table. 15.3.

656 Chapter 1 5

Table 15.3 Application of CRT in Pile Design (After De Ruiter and Beringen, 1979)

Item Sand Clay Legend

Unit Friction

Unit end bearing qt

Minimum of

,/; = 012MPa

/2 = CPT sleeve friction

/3 = <7r/300 (compression)

or

/3 = <?f/400 (tension)

minimum of (7 from

Fig. 15.17(b) andc

f s = t f cu, where

of = 1 in N.C. clay

= 0.5 in O.C. clay

q = N c*p c u

N =9

qc - cone resistance

below pile tip

q; = ultimate

resistance of pile

Ultimate Skin Load Qf in Cohesionless Soils

For the computation of skin load, £>, Schmertmann (1978) presents the following equation

— / A + f Ao , J C S Jc S (15.56c)

where K = fjf = correction factor for/c

fs = unit pile frictionfc = unit sleeve friction measured by the friction jacketz = depth to/c value considered from ground surfaced = pile diameter or width

A = pile-soil contact area per/c depth intervalL = embedded depth of pile.

When/c does not vary significantly with depth, Eq. (15.56c) can be written in a simplifiedform as

-Cf2 c (15.56d)

where fc is the average value within the depths specified. The correction factor K is given inFig. 15.18(a).

Ultimate Skin Load Qf for Piles in Clay Soil

For piles in clay Schmertmann gives the expression

(15.57)

where, of = ratio of pile to penetrometer sleeve friction,/ = average sleeve friction,A = pile to soil contact area.

Fig. 15.18(b) gives values of a'.


K values for steel pipe pile0 1.0 2.0 3.0

K values for concrete pile0 1.0 2.0

10

3 20

30

40

/

(^timber- 1-25 A'

1. Mechanical penetrometer2. Electrical penetrometer

(a)

Penetrometer sleeve friction, fc kg/cm

(b)

Figure 15.18 Penetrometer design curves (a) for pile side friction in sand(Schmertmann, 1978), and (b) for pile side friction in clay (Schmertmann, 1978)

Example 15.19

A concrete pile of 40 cm diameter is driven into a homogeneous mass of cohesionless soil. The pilecarries a safe load of 650 kN. A static cone penetration test conducted at the site indicates anaverage value of qc = 40 kg/cm2 along the pile and 120 kg/cm2 below the pile tip. Compute thelength of the pile with Fs = 2.5. (Fig. Ex. 15.19)

658 Chapter 15

Solution

From Eq. (15.5la)

qb (pile) = qp (cone)

Given, q = 120 kg/cm2, therefore,

qb=l2Q kg/cm2 = 120 x 100 = 12000 kN/m2

Per Section 15.12, qb is restricted to 11,000 kN/m2.Therefore,

314Qb =Abqb =— x0.42x 11000 = 1382 kN

Assume the length of the pile = L m

The average, qc = 40 kg/cm2

PerEq. (15.52a),

fs = —- kN/m2 = — = 20 kN/m2

Now, Qf = fsAs = 20 x 3.14 x 0.4 x L = 25.12L kN

Given Qa = 650 kN. With Fv = 2.5, Qu = 650 x 2.5 = 1625 kN.

Now,

1625-1382or L =

25.12= 9.67 m or say 10 m

The pile has to be driven to a depth of 10 m to carry a safe load of 650 kN with F =2.5.

I!"

qc - 40 kg/cm

qc= 120 kg/cm2

Qb

Figure Ex. 15.19


Example 15.20A concrete pile of size 0.4 x 0.4 m is driven to a depth of 12 m into medium dense sand. The watertable is close to the ground surface. Static cone penetration tests were carried out at this site byusing an electric cone penetrometer. The values of qc and fc as obtained from the test have beenplotted against depth and shown in Fig. Ex. 15.20. Determine the safe load on this pile with Fs = 2.5by Schmertmann's method (Section 15.21).

Solution

First determine the representative cone penetration value q by using Eq. (15.54)

Now from Fig. Ex. 15.20 and Eq (15.56a)

1 4d

_ 0.7(76 + 85) / 2 + 0.3(85 + 71) / 2 + 0.6(71 + 80) / 2

4x0.4

= 78 kg/cm2

= qd = the minimum value below the tip of pile within 4d depth = 71 kg/cm2.

qc, kg/cm2 fc, kg/cm2

,0 20 40 60 80 0 0.5 1.0

\Qu

Square concretepile 0.4 x 0.4 m

z= L= 12m

Figure Ex. 15.20

660 Chapter 15

FromEq. (15.56b)

_ 0.4x71 + 0.3(71 + 65)72 + 2.1x65 + 0.4(65 + 60)7 2

8x0.4

= 66 kg/cm2 = 660 t/m2 (metric)

From Eq. (15.54)

(78 + 71)72 + 66 ^ 700 t / m2 (metric)

Ultimate Base Load

Qb = qb Ab = qp Ab - 700 x 0.42 = 1 1 2 t (metric)

Frictional Load Qf

From Eq. (15.56d)

l-L

where K = correction factor from Fig. (15.18a) for electrical penetrometer.

L _ 12For — - T~ - 3v , K = 0.75 for concrete pile. It is now necessary to determine the average

sleeve friction fc between depths z = 0 and z - 8d, and z = 8d and z = L from the top of pile from/c

profile given in Fig. Ex. 15.20.

Q =0.75[|x0.34x 10x4x0.4x3.2 + 0.71x10x4x0.4x8.8]

= 0.75 [8.7 + 99.97] = 81.5 t (metric)

Q =Qh + Qf= 112 + 81.5= 193.5 t*^u *z-b - 7

Qh+Qr 1935Q = -2 L = _—L_ = 77.4 t (metric) = 759 kN•*" /7 r\ c s*\ C2.5 2.5

Example 15.21A concrete pile of section 0.4 x 0.4 m is driven into normally consolidated clay to a depth of 10 m.The water table is at ground level. A static cone penetration test (CPT) was conducted at the sitewith an electric cone penetrometer. Fig. Ex. 15.21 gives a profile of qc and/c values with respect todepth. Determine safe loads on the pile by the following methods:

(a) a-method (b) A-method, given: yb = 8.5 kN/m3 and (c) Schmertmann's method. Use afactor of safety of 2.5.

Solution

(a) a-methodThe a-method requires the undrained shear strength of the soil. Since this is not given, it has

to be determined by using the relation between qc and cu given in Eq. (9.14).


qc, kg/cm5 10 15

Square concretepile 0.4 x 0.4 m

Figure Ex. 15.21

cu jy by neglecting the overburden effect,YV k

where Nk = cone factor = 20.It is necessary to determine the average ~cu along the pile shaft and cb at the base level of the

pile. For this purpose find the corresponding^ (sleeve friction) values from Fig. Ex. 15.21.

- _ 1 + 1 6_c^Average qc along the shaft, Qc r <>•••> kg

Average of qc within a depth 3d above the base and d below the base of the pile (Refer toFig. 15.17a)

15 + 18.52= 17kg/cirr

o c

c = — = 0.43 kg/cm2 « 4 3 k N / m 2

" 20

662 Chapter 15

c , = — = 0.85kg/cm2 = 85kN/m 2 .b 20Ultimate Base Load, Qb

FromEq. (15.39)

Qb = 9cbAb = 9 x 85 x 0.402 = 122 kN.

Ultimate Friction Load, Qf

FromEq. (15.37)

Qf=<*uAs

From Fig. 15.15 for cu = 43 kN/m2, a = 70

Qf= 0.70 x 43 x 10 x 4 x 0.4 = 481.6 kN or say 482 kNQu = 122 + 482 = 604 kN

604Qa = _ = 241.6 kN or say 242 kN

jL,*j

(b) A-methodBase Load Qb

In this method the base load is the same as in (a) above. That isQb = 122 kN

Friction LoadFrom Fig. 15.17 A = 0.25 for L = 10 m (= 32.4 ft). From Eq. 15.40

q =1x10x8.5 = 42.5 kN/m 2H° 2

fs = 0.25(42.5 + 2 x 43) = 32 kN / m2

Q =fsAs = 32x10x4x0.4 = 512 kN

Q =122 + 512 = 634 kN*^u

Q = — = 254kN.a 2.5

(c) Schmertmann's MethodBase load Qb

Use Eq. (15.54) for determining the representative value for qp. Here, the minimum value forqc is at point O on the ^-profile in Fig. Ex. 15.21 which is the base level of the pile. Now qcl is theaverage qc at the base and O.ld below the base of the pile, that is,

q0+ge 16 + 18.5 2qcl = — - - - - = 1 7.25 kg / cm2

= The average of qc within a depth &d above the base level


q0+4k 16 + 11 , , _ . . 2= — = = 13.5 kg/cm2

2 2

(1725 + 16)72+13.5 1 C 1 . 2From Eq. (15.45), Vp = r = 15 kg7001 .

2

FromEq. (15.55a)

^(pile) = qp (cone) = 15 kg/cm2 « 1500 kN/m2

Qfc = 9ftAfc = 1500 x (0.4)2 = 240 kN

Friction Load Qf

Use Eq. (15.57)

where ex' = ratio of pile to penetrometer sleeve friction.

From Fig. Ex. 15.21 fc = 0 + L15 = 0.58 kg/cm2 « 58 kN/m2.

From Fig. 16.18b for/c = 58 kN/m2, a' = 0.70

Qf= 0.70 x 58 x 10 x 4 x 0.4 = 650 kN

Qu = 240 + 650 = 890 kN

Q =^^ = 356kNa 2.5

Note: The values given in the examples are only illustrative and not factual.

15.22 BEARING CAPACITY OF A SINGLE PILE BY LOAD TESTA pile load test is the most acceptable method to determine the load carrying capacity of a pile. Theload test may be carried out either on a driven pile or a cast-in-situ pile. Load tests may be madeeither on a single pile or a group of piles. Load tests on a pile group are very costly and may beundertaken only in very important projects.

Pile load tests on a single pile or a group of piles are conducted for the determination of

1. Vertical load bearing capacity,2. Uplift load capacity,

3. Lateral load capacity.

Generally load tests are made to determine the bearing capacity and to establish the load-settlement relationship under a compressive load. The other two types of tests may be carried outonly when piles are required to resist large uplift or lateral forces.

Usually pile foundations are designed with an estimated capacity which is determined from athorough study of the site conditions. At the beginning of construction, load tests are made for thepurpose of verifying the adequacy of the design capacity. If the test results show an inadequatefactor of safety or excessive settlement, the design must be revised before construction is under way.

Load tests may be carried out either on

1. A working pile or2. A test pile.

664 Chapter 15

A working pile is a pile driven or cast-in-situ along with the other piles to carry the loads fromthe superstructure. The maximum test load on such piles should not exceed one and a half times thedesign load.

A test pile is a pile which does not carry the loads coming from the structure. The maximumload that can be put on such piles may be about 2J/2 times the design load or the load imposed mustbe such as to give a total settlement not less than one-tenth the pile diameter.

Method of Carrying Out Vertical Pile Load Test

A vertical pile load test assembly is shown in Fig. 15.19(a). It consists of

1. An arrangement to take the reaction of the load applied on the pile head,2. A hydraulic jack of sufficient capacity to apply load on the pile head, and3. A set of three dial gauges to measure settlement of the pile head.

Load ApplicationA load test may be of two types:

1. Continuous load test.2. Cyclic load test.

In the case of a continuous load test, continuous increments of load are applied to the pilehead. Settlement of the pile head is recorded at each load level.

In the case of the cyclic load test, the load is raised to a particular level, then reduced to zero,again raised to a higher level and reduced to zero. Settlements are recorded at each increment ordecrement of load. Cyclic load tests help to separate frictional load from point load.

The total elastic recovery or settlement Se, is due to

1. The total plastic recovery of the pile material,2. Elastic recovery of the soil at the tip of the pile, Sg

The total settlement S due to any load can be separated into elastic and plastic settlements bycarrying out cyclic load tests as shown in Fig. 15.19(b).

A pile loaded to Ql gives a total settlement Sr When this load is reduced to zero, there is anelastic recovery which is equal to Sel. This elastic recovery is due to the elastic compression of the pilematerial and the soil. The net settlement or plastic compression is S r The pile is loaded again fromzero to the next higher load Q2 and reduced to zero thereafter. The corresponding settlements may befound as before. The method of loading and unloading may be repeated as before.

Allowable Load from Single Pile Load Test Data

There are many methods by which allowable loads on a single pile may be determined by makinguse of load test data. If the ultimate load can be determined from load-settlement curves, allowableloads are found by dividing the ultimate load by a suitable factor of safety which varies from 2 to 3.A factor of safety of 2.5 is normally recommended. A few of the methods that are useful for thedetermination of ultimate or allowable loads on a single pile are given below:

1. The ultimate load, Qu, can be determined as the abscissa of the point where the curved partof the load-settlement curve changes to a falling straight line. Fig. 15.20(a).

2. Qu is the abscissa of the point of intersection of the initial and final tangents of theload-settlement curve, Fig. 15.20(b).

3. The allowable load Q is 50 percent of the ultimate load at which the total settlementamounts to one-tenth of the diameter of the pile for uniform diameter piles.


Anchor rod

Anchor pile

(a)

//A

I

M^\t

A C

V T

Pile 1 = before load2 = after load3 = after elastic recovery

(b)

Figure 15.19 (a) Vertical pile load test assembly, and (b) elastic compression atthe base of the pile

4. The allowable load Qa is sometimes taken as equal to two-thirds of the load which causes atotal settlement of 12 mm.

5. The allowable load Qa is sometimes taken as equal to two-thirds of the load which causes anet (plastic) settlement of 6 mm.

If pile groups are loaded to failure, the ultimate load of the group, Q u, may be found by any oneof the first two methods mentioned above for single piles. However, if the groups are subjected to only

666 Chapter 1 5

Q

(a) (b)

Figure 15.20 Determination of ultimate load from load-settlement curves

one and a half-times the design load of the group, the allowable load on the group cannot be found onthe basis of 12 or 6 mm settlement criteria applicable to single piles. In the case of a group with pilesspaced at less than 6 to 8 times the pile diameter, the stress interaction of the adjacent piles affects thesettlement considerably. The settlement criteria applicable to pile groups should be the same as thatapplicable to shallow foundations at design loads.

15.23 PILE BEARING CAPACITY FROM DYNAMIC PILE DRIVINGFORMULASThe resistance offered by a soil to penetration of a pile during driving gives an indication of itsbearing capacity. Qualitatively speaking, a pile which meets greater resistance during driving iscapable of carrying a greater load. A number of dynamic formulae have been developed whichequate pile capacity in terms of driving energy.

The basis of all these formulae is the simple energy relationship which may be stated by thefollowing equation. (Fig. 15.21).

Wh = Qus

or Qu=— (15.58)5

where W - weight of the driving hammer

h = height of fall of hammer

Wh = energy of hammer blow

Q = ultimate resistance to penetration

s = pile penetration under one hammer blow

Qus = resisting energy of the pile

Hiley FormulaEquation (15.58) holds only if the system is 100 percent efficient. Since the driving of a pileinvolves many losses, the energy of the system may be written as


$x Hammer

t,h\ '

J/Pile^kcap

s = penetration ^ |fof pile ^

r x x x

r

3

x — Wp = weight of pile

f tQu

Figure 15.21 Basic energy relationship

Energy input = Energy used + Energy losses

or Energy used = Energy input - Energy losses.

The expressions for the various energy terms used are

1 . Energy used = Qus,

2. Energy input = T]hWh, where r\h is the efficiency of the hammer.

3. The energy losses are due to the following:(i) The energy loss E{ due to the elastic compressions of the pile cap, pile material and

the soil surrounding the pile. The expression for El may be written as

where Cj = elastic compression of the pile capc2 = elastic compression of the pilec3 = elastic compression of the soil.

(ii) The energy loss E2 due to the interaction of the pile hammer system (impact of twobodies). The expression for E2 may be written as

i-cr2

£7 = WhW -2 P W+W

where W = weight of pileC = coefficient of restitution.

668 Chapter 15

Substituting the various expressions in the energy equation and simplifying, we have

^ rj.Wh l + C2R

TTF (15'59>w

where R - ——W

Equation (15.59) is called the Hiley formula. The allowable load Qu may be obtained bydividing Qu by a suitable factor of safety.

If the pile tip rests on rock or relatively impenetrable material, Eq. (15.59) is not valid. Chellis(1961) suggests for this condition that the use of W 12 instead of W may be more correct. Thevarious coefficients used in the Eq. (15.59) are as given below:

1. Elastic compression c{ of cap and pile head

Pile Material Range of Driving Range of c1

Stress kg/cm2

Precast concrete pile withpacking inside cap 30-150 0.12-0-50

Timber pile without cap 30-150 0.05-0.20Steel //-pile 30-150 0.04-0.16

2. Elastic compression c2 of pile.This may be computed using the equation

c -~AE

where L = embedded length of the pile,A = average cross-sectional area of the pile,E = Young's modulus.

3. Elastic compression c3 of soil.The average value of c3 may be taken as 0. 1 (the value ranges from 0.0 for hard soil to 0.2 for

resilient soils).

4. Pile-hammer efficiency

Hammer Type r\h

DropSingle actingDouble actingDiesel

1.000.75-0.850.851.00

5. Coefficient of restitution Cr

Material Cr

Wood pile 0.25Compact wood cushion on steel pile 0.32Cast iron hammer on concrete pile without cap 0.40Cast iron hammer on steel pipe without cushion 0.55


Engineering News Record (ENR) Formula

The general form of the Engineering News Record Formula for the allowable load Qa may beobtained from Eq. (15.59) by putting

r]h = 1 and Cr = 1 and a factor of safety equal to 6. The formula proposed by A.M. Wellington,editor of the Engineering News, in 1886, is

Wh .(15-60)

where Qa = allowable load in kg,

W = weight of hammer in kg,

h = height of fall of hammer in cm,

s - final penetration in cm per blow (which is termed as set). The set is taken as theaverage penetration per blow for the last 5 blows of a drop hammer or 20 blows of asteam hammer,

C = empirical constant,

= 2.5 cm for a drop hammer,

= 0.25 cm for single and double acting hammers.

The equations for the various types of hammers may be written as:

1 . Drop hammer

Wh

2. Single-acting hammer

m(15-62)

3. Double-acting hammer

(W + ap)

a = effective area of the piston in sq. cm,

p = mean effective steam pressure in kg/cm2.

Comments on the Use of Dynamic Formulae

1. Detailed investigations carried out by Vesic (1967) on deep foundations in granular soilsindicate that the Engineering News Record Formula applicable to drop hammers,Eq. (15.61), gives pile loads as low as 44 % of the actual loads. In order to obtain betteragreement between the one computed and observed loads, Vesic suggests the followingvalues for the coefficient C in Eq. (15.60).

For steel pipe piles, C = 1 cm.

For precast concrete piles C = 1.5 cm.

2. The tests carried out by Vesic in granular soils indicate that Hiley's formula does not giveconsistent results. The values computed from Eq. (15.59) are sometimes higher andsometimes lower than the observed values.

670 Chapter 15

3. Dynamic formulae in general have limited value in pile foundation work mainly because thedynamic resistance of soil does not represent the static resistance, and because often the resultsobtained from the use of dynamic equations are of questionable dependability. However,engineers prefer to use the Engineering News Record Formula because of its simplicity.

4. Dynamic formulae could be used with more confidence in freely draining materials such ascoarse sand. If the pile is driven to saturated loose fine sand and silt, there is everypossibility of development of liquefaction which reduces the bearing capacity of the pile.

5. Dynamic formulae are not recommended for computing allowable loads of piles driveninto cohesive soils. In cohesive soils, the resistance to driving increases through the suddenincrease in stress in pore water and decreases because of the decreased value of the internalfriction between soil and pile because of pore water. These two oppositely directed forcesdo not lend themselves to analytical treatment and as such the dynamic penetrationresistance to pile driving has no relationship to static bearing capacity.

There is another effect of pile driving in cohesive soils. During driving the soil becomesremolded and the shear strength of the soil is reduced considerably. Though there will be aregaining of shear strength after a lapse of some days after the driving operation, this will not bereflected in the resistance value obtained from the dynamic formulae.

Example 15.22A 40 x 40 cm reinforced concrete pile 20 m long is driven through loose sand and then into densegravel to a final set of 3 mm/blow, using a 30 kN single-acting hammer with a stroke of 1.5 m.Determine the ultimate driving resistance of the pile if it is fitted with a helmet, plastic dolly and 50mm packing on the top of the pile. The weight of the helmet and dolly is 4 kN. The other details are:weight of pile = 74 kN; weight of hammer = 30 kN; pile hammer efficiency rjh = 0.80 andcoefficient of restitution Cr = 0.40.

Use the Hiley formula. The sum of the elastic compression C is

C = cl +c2 +c3 = 19.6 mm.

Solution

Hiley Formula

Use Eq. (15.59)

„ ^ *" s + C \ + R

Wwhere n. = 0.80, W = 30 kN, h = 1.5 m, R = —p- = ——- = 2.6, C = 0.40, s = 0.30 cm.h W 30

Substituting we have,

0.8x30x150 l + 0.42x2.6

0.3 + 156/2 l + 2.6 =2813x0.393=1105kN

15.24 BEARING CAPACITY OF PILES FOUNDED ON A ROCKY BEDPiles are at times required to be driven through weak layers of soil until the tips meet a hard stratafor bearing. If the bearing strata happens to be rock, the piles are to be driven to refusal in order toobtain the maximum carrying capacity from the piles. If the rock is strong at its surface, the pile will


refuse further driving at a negligible penetration. In such cases the carrying capacity of the piles isgoverned by the strength of the pile shaft regarded as a column as shown in Fig. 15.6(a). If the soilmass through which the piles are driven happens to be stiff clays or sands, the piles can be regardedas being supported on all sides from buckling as a strut. In such cases, the carrying capacity of a pileis calculated from the safe load on the material of the pile at the point of minimum cross-section. Inpractice, it is necessary to limit the safe load on piles regarded as short columns because of thelikely deviations from the vertical and the possibility of damage to the pile during driving.

If piles are driven to weak rocks, working loads as determined by the available stress on thematerial of the pile shaft may not be possible. In such cases the frictional resistance developed overthe penetration into the rock and the end bearing resistance are required to be calculated. Tomlinson(1986) suggests an equation for computing the end bearing resistance of piles resting on rockystrata as

<7u=2AVV (15.64)

where N^ = tan2 (45 + 0/2),qur - unconfmed compressive strength of the rock.

Boring of a hole in rocky strata for constructing bored piles may weaken the bearing strata ofsome types of rock. In such cases low values of skin friction should be used and normally may notbe more than 20 kN/m2 (Tomlinson, 1986) when the boring is through friable chalk or mud stone.In the case of moderately weak to strong rocks where it is possible to obtain core samples forunconfmed compression tests, the end bearing resistance can be calculated by making use ofEq. (15.64).

15.25 UPLIFT RESISTANCE OF PILESPiles are also used to resist uplift loads. Piles used for this purpose are called tension piles, upliftpiles or anchor piles. Uplift forces are developed due to hydrostatic pressure or overturningmoments as shown in Fig. 15.22.

Figure 15.22 shows a straight edged pile subjected to uplift force. The equation for the upliftforce PU[ may be written as

(15.65)

where, Pul = uplift capacity of pile,W = weight of pile,/ = unit resisting forceAs = effective area of the embedded length of pile.

Uplift Resistance of Pile in ClayFor piles embedded in clay, Eq. (15.65) may written as

(15.66)

where, cu = average undrained shear strength of clay along the pile shaft,a, = adhesion factor (= cfl/cM),ca = average adhesion.

Figure 15.23 gives the relationship between a and cu based on pull out test results as collectedby Sowa (1970). As per Sowa, the values of ca agree reasonably well with the values for pilessubjected to compression loadings.

672 Chapter 1 5

fr

Figure 15.22 Single pile subjected to uplift

1.25

0.75

0.50

0.25

Average curve for concrete pile

Average curve for all pile

25 50 75 100 125Undrained shear strength cu, kN/m

150

Figure 15.23 Relationship between adhesion factor a and undrained shear strengthc (Source: Poulos and Davis, 1980)

Uplift Resistance of Pile in SandAdequate confirmatory data are not available for evaluating the uplift resistance of piles embeddedin cohesionless soils. Ireland (1957) reports that the average skin friction for piles undercompression loading and uplift loading are equal, but data collected by Sowa (1970) and Downsand Chieurzzi (1966) indicate lower values for upward loading as compared to downward loadingespecially for cast-in-situ piles. Poulos and Davis (1980) suggest that the skin friction of upwardloading may be taken as two-thirds of the calculated shaft resistance for downward loading.

A safety factor of 3 is normally assumed for calculating the safe uplift load for both piles inclay and sand.


Example 15.23A reinforced concrete pile 30 ft long and 15 in. in diameter is embedded in a saturated clay of verystiff consistency. Laboratory tests on samples of undisturbed soil gave an average undrainedcohesive strength cu = 2500 lb/ft2. Determine the net pullout capacity and the allowable pullout loadwith Fs = 3.

Solution

Given: L = 30 ft, d = 15 in. diameter, cu = 2500 lb/ft2, Fs = 3.

From Fig. 15.23 cjcu = 0.41 for cu = 2500 x 0.0479 « 120 kN/m2 for concrete pile.

From Eq. (15.66)

where a = c/c= 0.41, c = 2500 lb/ft2

A =3.14x — x 30 = 117.75 ft2

12

Substituting

„ , , 0.41x2500x117.75 ,„„„, , .P,(net) = - = 120.69 kipsul 1000 P

Pul (allowed) = ^40 kips

Example 15.24Refer to Ex. 15.23. If the pile is embedded in medium dense sand, determine the net pulloutcapacity and the net allowable pullout load with Fs = 3.

Given: L = 30 ft, 0 = 38°, ~Ks = 1.5, and 8 = 25°, 7 (average) = 1.10 lb/ft3.The water table is at great depth. Refer to Section 15.25.

Solution

Downward skin resistance Qf

where q' =-x 30x1 10 = 1650 lb/ft2"o

A =3.14x1.25x30 = 117.75 ft2

N 1650 x 1.5 tan25°x 117.75<2/-(down) = = 136 kipsf 1000

Based on the recommendations of Poulos and Davis (1980)

2 2= — <2/(down) = — x 136 = 91 kips

674 Chapter 1 5

kips

PART B—PILE GROUP

15.26 NUMBER AND SPACING OF PILES IN A GROUPVery rarely are structures founded on single piles. Normally, there will be a minimum of three pilesunder a column or a foundation element because of alignment problems and inadvertenteccentricities. The spacing of piles in a group depends upon many factors such as

Pile cap

(a) Single pile

Isobar of asingle pile

Isobar ofa group

Highlystressedzone

(b) Group of piles closely spaced

Q

i r Pile cap

(c) Group of piles with piles far apart

Figure 15.24 Pressure isobars of (a) single pile, (b) group of piles, closely spaced,and (c) group of piles with piles far apart.


1. overlapping of stresses of adjacent piles,

2. cost of foundation,

3. efficiency of the pile group.

The pressure isobars of a single pile with load Q acting on the top are shown in Fig. 15.24(a).When piles are placed in a group, there is a possibility the pressure isobars of adjacent piles willoverlap each other as shown in Fig. 15.24(b). The soil is highly stressed in the zones of overlappingof pressures. With sufficient overlap, either the soil will fail or the pile group will settle excessivelysince the combined pressure bulb extends to a considerable depth below the base of the piles. It ispossible to avoid overlap by installing the piles further apart as shown in Fig. 15.24(c). Largespacings are not recommended sometimes, since this would result in a larger pile cap which wouldincrease the cost of the foundation.

The spacing of piles depends upon the method of installing the piles and the type of soil. Thepiles can be driven piles or cast-in-situ piles. When the piles are driven there will be greateroverlapping of stresses due to the displacement of soil. If the displacement of soil compacts the soilin between the piles as in the case of loose sandy soils, the piles may be placed at closer intervals.

—

10 pile 11 pile 12 pile

Figure 15.25 Typical arrangements of piles in groups

676 Chapter 15

But if the piles are driven into saturated clay or silty soils, the displaced soil will not compact thesoil between the piles. As a result the soil between the piles may move upwards and in this processlift the pile cap. Greater spacing between piles is required in soils of this type to avoid lifting ofpiles. When piles are cast-in-situ, the soils adjacent to the piles are not stressed to that extent and assuch smaller spacings are permitted.

Generally, the spacing for point bearing piles, such as piles founded on rock, can be muchless than for friction piles since the high-point-bearing stresses and the superposition effect ofoverlap of the point stresses will most likely not overstress the underlying material nor causeexcessive settlements.

The minimum allowable spacing of piles is usually stipulated in building codes. The spacingsfor straight uniform diameter piles may vary from 2 to 6 times the diameter of the shaft. For frictionpiles, the minimum spacing recommended is 3d where d is the diameter of the pile. For end bearingpiles passing through relatively compressible strata, the spacing of piles shall not be less than 2.5d.For end bearing piles passing through compressible strata and resting in stiff clay, the spacing maybe increased to 3.5d. For compaction piles, the spacing may be Id. Typical arrangements of piles ingroups are shown in Fig. 15.25.

15.27 PILE GROUP EFFICIENCYThe spacing of piles is usually predetermined by practical and economical considerations. Thedesign of a pile foundation subjected to vertical loads consists of

1. The determination of the ultimate load bearing capacity of the group Q u.2, Determination of the settlement of the group, S , under an allowable load Q .

a o

The ultimate load of the group is generally different from the sum of the ultimate loads ofindividual piles Qu.

The factor

QguE = ( '1^A7^

is called group efficiency which depends on parameters such as type of soil in which the piles areembedded, method of installation of piles i.e. either driven or cast-in-situ piles, and spacing ofpiles.

There is no acceptable "efficiency formula" for group bearing capacity. There are a fewformulae such as the Converse-Labarre formula that are sometimes used by engineers. Theseformulae are empirical and give efficiency factors less than unity. But when piles are installed insand, efficiency factors greater than unity can be obtained as shown by Vesic (1967) by hisexperimental investigation on groups of piles in sand. There is not sufficient experimentalevidence to determine group efficiency for piles embedded in clay soils.

Efficiency of Pile Groups in SandVesic (1967) carried out tests on 4 and 9 pile groups driven into sand under controlled conditions.Piles with spacings 2, 3,4, and 6 times the diameter were used in the tests. The tests were conductedin homogeneous, medium dense sand. His findings are given in Fig. 15.26. The figure gives thefollowing:

1. The efficiencies of 4 and 9 pile groups when the pile caps do not rest on the surface.2. The efficiencies of 4 and 9 pile groups when the pile caps rest on the surface.


3.0

677

2.5

tq

56<uQ.

§

E

1.0 _DQ

/ // /

n

i-

1. Point efficiency—average of all tests

2. 4 pile group—total efficiency

3. 9 pile group—total efficiency

4. 9 pile group—total efficiency with cap

5. 4 pile group—total efficiency with cap

6. 4 pile group—skin efficiency

7. 9 pile group—skin efficiency

1 2 3 4 5 6 7Pile spacing in diameters

Figure 15.26 Efficiency of pile groups in sand (Vesic, 1967)

3. The skin efficiency of 4 and 9 pile groups.4. The average point efficiency of all the pile groups.

It may be mentioned here that a pile group with the pile cap resting on the surface takes moreload than one with free standing piles above the surface. In the former case, a part of the load istaken by the soil directly under the cap and the rest is taken by the piles. The pile cap behaves thesame way as a shallow foundation of the same size. Though the percentage of load taken by thegroup is quite considerable, building codes have not so far considered the contribution made by thecap.

It may be seen from the Fig. 15.26 that the overall efficiency of a four pile group with a capresting on the surface increases to a maximum of about 1.7 at pile spacings of 3 to 4 pile diameters,becoming somewhat lower with a further increase in spacing. A sizable part of the increasedbearing capacity comes from the caps. If the loads transmitted by the caps are reduced, the groupefficiency drops to a maximum of about 1.3.

Very similar results are indicated from tests with 9 pile groups. Since the tests in this casewere carried out only up to a spacing of 3 pile diameters, the full picture of the curve is notavailable. However, it may be seen that the contribution of the cap for the bearing capacity isrelatively smaller.

Vesic measured the skin loads of all the piles. The skin efficiencies for both the 4 and 9-pilegroups indicate an increasing trend. For the 4-pile group the efficiency increases from about 1.8 at2 pile diameters to a maximum of about 3 at 5 pile diameters and beyond. In contrast to this, theaverage point load efficiency for the groups is about 1.01. Vesic showed for the first time that the

678 Chapter 15

increasing bearing capacity of a pile group for piles driven in sand comes primarily from anincrease in skin loads. The point loads seem to be virtually unaffected by group action.

Pile Group Efficiency Equation

There are many pile group equations. These equations are to be used very cautiously, and may inmany cases be no better than a good guess. The Converse-Labarre Formula is one of the mostwidely used group-efficiency equations which is expressed as

6(n - \}m + (m- l)n(15'68>

where m = number of columns of piles in a group,n = number of rows,6 = lan~l( d/s) in degrees,d = diameter of pile,s = spacing of piles center to center.

15.28 VERTICAL BEARING CAPACITY OF PILE GROUPSEMBEDDED IN SANDS AND GRAVELSDriven piles. If piles are driven into loose sands and gravel, the soil around the piles to a radius ofat least three times the pile diameter is compacted. When piles are driven in a group at closespacing, the soil around and between them becomes highly compacted. When the group is loaded,the piles and the soil between them move together as a unit. Thus, the pile group acts as a pierfoundation having a base area equal to the gross plan area contained by the piles. The efficiency ofthe pile group will be greater than unity as explained earlier. It is normally assumed that theefficiency falls to unity when the spacing is increased to five or six diameters. Since presentknowledge is not sufficient to evaluate the efficiency for different spacing of piles, it is conservativeto assume an efficiency factor of unity for all practical purposes. We may, therefore, write

Q =nQ (15.69)^-gu *^u ^ '

where n - the number of piles in the group.The procedure explained above is not applicable if the pile tips rest on compressible soil such as

silts or clays. When the pile tips rest on compressible soils, the stresses transferred to the compressiblesoils from the pile group might result in over-stressing or extensive consolidation. The carrying capacityof pile groups under these conditions is governed by the shear strength and compressibility of the soil,rather than by the 'efficiency'' of the group within the sand or gravel stratum.

Bored Pile Groups In Sand And Gravel

Bored piles are cast-in-situ concrete piles. The method of installation involves

1 . Boring a hole of the required diameter and depth,2. Pouring in concrete.

There will always be a general loosening of the soil during boring and then too when theboring has to be done below the water table. Though bentonite slurry (sometimes called as drillingmud) is used for stabilizing the sides and bottom of the bores, loosening of the soil cannot beavoided. Cleaning of the bottom of the bore hole prior to concreting is always a problem which willnever be achieved quite satisfactorily. Since bored piles do not compact the soil between the piles,


'TirTTT

Perimeter Pn

Figure 15.27 Block failure of a pile group in clay soil

the efficiency factor will never be greater than unity. However, for all practical purposes, theefficiency may be taken as unity.

Pile Groups In Cohesive Soils

The effect of driving piles into cohesive soils (clays and silts) is very different from that ofcohesionless soils. It has already been explained that when piles are driven into clay soils, particularlywhen the soil is soft and sensitive, there will be considerable remolding of the soil. Besides there willbe heaving of the soil between the piles since compaction during driving cannot be achieved in soils ofsuch low permeability. There is every possibility of lifting of the pile during this process of heaving ofthe soil. Bored piles are, therefore, preferred to driven piles in cohesive soils. In case driven piles areto be used, the following steps should be favored:

1. Piles should be spaced at greater distances apart.2. Piles should be driven from the center of the group towards the edges, and3. The rate of driving of each pile should be adjusted as to minimize the development of pore

water pressure.

Experimental results have indicated that when a pile group installed in cohesive soils isloaded, it may fail by any one of the following ways:

1. May fail as a block (called block failure).2. Individual piles in the group may fail.

680 Chapter 15

When piles are spaced at closer intervals, the soil contained between the piles movedownward with the piles and at failure, piles and soil move together to give the typical 'blockfailure'. Normally this type of failure occurs when piles are placed within 2 to 3 pile diameters. Forwider spacings, the piles fail individually. The efficiency ratio is less than unity at closer spacingsand may reach unity at a spacing of about 8 diameters.

The equation for block failure may be written as (Fig. 15.27).

Qgu=cNcAg

+pg^ (15.70)

where c = cohesive strength of clay beneath the pile group,c = average cohesive strength of clay around the group,L = length of pile,

P = perimeter of pile group,o

A = sectional area of group,0

Nc = bearing capacity factor which may be assumed as 9 for deep foundations.The bearing capacity of a pile group on the basis of individual pile failure may be written as

Q =nQ (15.71)*^gu ^-u ^ '

where n = number of piles in the group,

Qu = bearing capacity of an individual pile.The bearing capacity of a pile group is normally taken as the smaller of the two given by Eqs.

(15.70) and (15.71).

Example 1 5.25A group of 9 piles with 3 piles in a row was driven into a soft clay extending from ground level toa great depth. The diameter and the length of the piles were 30 cm and 10m respectively. Theunconfmed compressive strength of the clay is 70 kPa. If the piles were placed 90 cm center tocenter, compute the allowable load on the pile group on the basis of a shear failure criterion for afactor of safety of 2.5.

Solution

The allowable load on the group is to be calculated for two conditions: (a) block failure and (b)individual pile failure. The least of the two gives the allowable load on the group.

(a) Block failure (Fig. 15.27). Use Eq. (15.70),

Q =cNA+PLc where N = 9, c = c = 70/2 = 35 kN/m2^-gU c R g C

A =2.1x2.1 = 4.4m2 , P = 4x 2.1 = 8.4 m, L = 10 m

Q =35x9x4.4 + 8.4x10x35 = 4326 kN, Q =^-gu ^-a 25

(b) Individual pile failure

Qu=Qb + Qf =<7^A + acAv . Assume a= 1.

Now, qb = cNc = 35 x 9 = 315 kN/m2, Ab = 0.07 m2,


As = 3.14x0.3x10 = 9.42m2

Substituting, Qu = 315 x 0.07 +1 x 35 x 9.42 = 352 kN

3168Qgu = «GU = 9 x 352 = 3168 kN> Qa = ~^~ = 1267 kN

The allowable load is 1267 kN.

15.29 SETTLEMENT OF PILES AND PILE GROUPS IN SANDS ANDGRAVELSNormally it is not necessary to compute the settlement of a single pile as this settlement under aworking load will be within the tolerable limits. However, settlement analysis of a pile group is verymuch essential. The total settlement analysis of a pile group does not bear any relationship with thatof a single pile since in a group the settlement is very much affected due to the interaction stressesbetween piles and the stressed zone below the tips of piles.

Settlement analysis of single piles by Poulos and Davis (1980) indicates that immediatesettlement contributes the major part of the final settlement (which includes the consolidationsettlement for saturated clay soils) even for piles in clay. As far as piles in sand is concerned, theimmediate settlement is almost equal to the final settlement.

However, it may be noted here that consolidation settlement becomes more important for pilegroups in saturated clay soils.

Immediate settlement of a single pile may be computed by making use of semi-empiricalmethods. The method as suggested by Vesic (1977) has been discussed here.

In recent years, with the advent of computers, more sophisticated methods of analysis havebeen developed to predict the settlement and load distribution in a single pile. The following threemethods are often used.

1. 'Load transfer' method which is also called as the 't-z' method.

2. Elastic method based on Mindlin's (1936) equations for the effects of subsurface loadingswithin a semi-infinite mass.

3. The finite element method.This chapter discusses only the 't-z' method. The analysis of settlement by the elastic method

is quite complicated and is beyond the scope of this book. Poulos and Davis, (1980) have discussedthis procedure in detail. The finite element method of analysis of a single pile axially loaded hasbeen discussed by many investigators such as Ellison et al., (1971), Desai (1974), Balaam et al.,(1975), etc. The finite element approach is a generalization of the elastic approach. The power ofthis method lies in its capability to model complicated conditions and to represent non-linear stress/strain behavior of the soil over the whole zone of the soil modelled. Use of computer programs isessential and the method is more suited to research or investigation of particularly complexproblems than to general design.

Present knowledge is not sufficient to evaluate the settlements of piles and pile groups. Formost engineering structures, the loads to be applied to a pile group will be governed byconsideration of consolidation settlement rather than by bearing capacity of the group divided byan arbitrary factor of safety of 2 or 3. It has been found from field observation that the settlement ofa pile group is many times the settlement of a single pile at the corresponding working load. Thesettlement of a group is affected by the shape and size of the group, length of piles, method ofinstallation of piles and possibly many other factors.

682 Chapter 15

Semi-Empirical Formulas and Curves

Vesic (1977) proposed an equation to determine the settlement of a single pile. The equation hasbeen developed on the basis of experimental results he obtained from tests on piles. Tests on piles ofdiameters ranging from 2 to 18 inches were carried out in sands of different relative densities. Testswere carried out on driven piles, jacked piles, and bored piles (jacked piles are those that are pushedinto the ground by using a jack). The equation for total settlement of a single pile may be expressedas

c - c . c (] s. 79\O — O t O r \ L J . l £ )

where S = total settlement,S = settlement of the pile tip,S, = settlement due to the deformation of the pile shaft.

The equation for Sp is

5 _ = - ^ (15.73)

The equation for 5, is

-^ (15.74)

where Q = point load,d = diameter of the pile at the base,

q u - ultimate point resistance per unit area,Dr = relative density of the sand,

Cw = settlement coefficient, = 0.04 for driven piles= 0.05 for jacked piles= 0.18 for bored piles,

Q, = friction load,L = pile length,A = cross-sectional area of the pile,E = modulus of deformation of the pile shaft,a, — coefficient which depends on the distribution of skin friction along the shaft and can

be taken equal to 0.6.Settlement of piles cannot be predicted accurately by making use of equations such as the

ones given here. One should use such equations with caution. It is better to rely on load tests forpiles in sands.

Settlement of Pile Groups in SandThe relation between the settlement of a group and a single pile at corresponding working loadsmay be expressed as

(15.75)

where F = group settlement factor,S = settlement of group,5 = settlement of a single pile.


s 10

ex

2O

10 20 30 40 50Relative width of the group Bid

60

Figure 15.28 Curve showing the relationship between group settlement ratio andrelative widths of pile groups in sand (Vesic, 1967)

16

*: 12

O z6 9 12 15

Width of pile group (m)18 21

Figure 15.29 Curve showing relationship between F and pile group width(Skempton, et al., 1953)

Vesic (1967) obtained the curve given in Fig. 15.28 by plotting F against Bid where d is thediameter of the pile and B, the distance between the center to center of the outer piles in the group(only square pile groups are considered). It should be remembered here that the curve is based onthe results obtained from tests on groups of piles embedded in medium dense sand. It is possiblethat groups in much looser or much denser deposits might give somewhat different behavior. Thegroup settlement ratio is very likely be affected by the ratio of the pile point settlement S to totalpile settlement.

Skempton et al., (1953) published curves relating F with the width of pile groups as shownin Fig. 15.29. These curves can be taken as applying to driven or bored piles. Since the abscissa forthe curve in Fig. 15.29 is not expressed as a ratio, this curve cannot directly be compared withVesic's curve given in Fig. 15.28. According to Fig. 15.29 a pile group 3 m wide would settle 5times that of a single test pile.

t-z MethodConsider a floating vertical pile of length L and diameter d subjected to a vertical load Q (Fig.15.30). This load will be transferred to the surrounding and underlying soil layers as described in

684 Chapter 15

Section 15.7. The pile load will be carried partly by skin friction (which will be mobilized throughthe increasing settlement on the mantle surface and the compression of the pile shaft) and partlythrough the pile tip, in the form of point resistance as can be seen from Fig. 15.30. Thus the load istaken as the sum of these components. The distribution of the point resistance is usually consideredas uniform; however, the distribution of the mantle friction depends on many factors. Load transferin pile-soil system is a very complex phenomenon involving a number of parameters which aredifficult to evaluate in numerical terms. Yet, some numerical assessment of load transfercharacteristics of a pile soil system is essential for the rational design of pile foundation.

The objective of a load transfer analysis is to obtain a load-settlement curve. The basicproblem of load transfer is shown in Fig. 15.30. The following are to be determined:

1. The vertical movements s, of the pile cross-sections at any depth z under loads acting on thetop,

2. The corresponding pile load Q7 at depth z acting on the pile section,

3. The vertical movement of the base of the pile and the corresponding point stress.

The mobilization of skin shear stress rat any depth z, from the ground surface depends on thevertical movement of the pile cross section at that level. The relationship between the two may belinear or non-linear. The shear stress T, reaches the maximum value, r,, at that section when thevertical movement of the pile section is adequate. It is, therefore, essential to construct (i - s)curves at various depths z as required.

There will be settlement of the tip of pile after the full mobilization of skin friction. Themovement of the tip, sh, may be assumed to be linear with the point pressure q . When themovement of the tip is adequate, the point pressure reaches the maximum pressure qb (ultimate basepressure). In order to solve the load-transfer problem, it is esential to construct a (q - s) curve.

(r - s) and (qp - s) curvesCoyle and Reese (1966) proposed a set of average curves of load transfer based on laboratory testpiles and instrumented field piles. These curves are limited to the case of steel-pipe friction piles inclay with an embedded depth not exceeding 100 ft. Coyle and Sulaiman (1967) have also givenload transfer curves for piles in sand. These curves are meant for specific cases and therefore meantto solve specific problems and as such this approach cannot be considered as a general case.

Verbrugge (1981) proposed an elastic-plastic model for the (r- s) and (q - s) curves basedon CPT results. The slopes of the elastic portion of the curves given are

r 0.22£(15.76)

s 2R

V 3.125E

2R(15.77)

The value of elastic modulus Ev of cohesionless soils may be obtained by the followingexpressions (Verbrugge);

for bored piles Ef = (36 + 2.2 qc) kg/cm2 (15.78)

for driven piles £\. = 3(36 + 2.2 qc) kg/cm2; (15.79)

where qc - point resistance of static cone penetrometer in kg/cm2.

The relationship recommended for E^ is for qc > 4 kg/cm2. The maximum values of T, and !„for the plastic portion of (r - s) curves are given in Table 15.4 and 15.5 for cohesionless andcohesive soils respectively. In Eqs (15.76) and (15.77) the value of E^ can be obtained by any one of


T - s curve

443

n - 3

n-2

n-1

^p(max)

qp - s curve

Qn-2

A^nrr

. 1

Pile subjected to Load transfer mechanismvertical road

Figure 15.30 t-z method of analysis of pile load-settlement relationship

the known methods. The maximum value of q (qb) may be obtained by any one of the knownmethods such as

1. From the relationship

qb = q'oN for cohesionless soils

qb = 9cu for cohesive soils.

Table 15.4 Recommended maximum skin shear stress rmax for piles incohesionless soils (After Verbrugge, 1981)

Tmax kN/m2 Pile type

0.011 qc Driven concrete piles

0.009 qc Driven steel piles

0.005 qc Bored concrete piles

Limiting values

T =80 kN/m2 for bored piles

T =120 kN/m2 for driven piles

686 Chapter 1 5

Table 15.5 Recommended maximum skin shear stress T „ for piles in cohesivemaxsoils (After Verbrugge, 1981)

[Values recommended are for Dutch cone penetrometer]

Type of pile

Driven

Bored

Material

Concrete

Steel

Concrete

Steel

Range of qc, kN/m 2

qc < 375375 < qc < 4500

4500 < qc

qc < 450

450<4 c < 1500

1500«?c.

qc<600

600 <qc< 4500

4500 < qc

qc< 500

500 <<? c < 1500

1500«?c.

max

0.053 qc

18 + 0.006 qc

0.01 qc

0.033 qc

15

0.01 qc

0.037 qc

18 + 0.006 qc

0.01 qc

0.03 <?c

15

0.01 qc

2. From static cone penetration test results

3. From pressuremeter test results

Method of obtaining load-settlement curve (Fig. 15.30)

The approximate load-settlement curve is obtained point by point in the following manner:

1. Divide the pile into any convenient segments (possibly 10 for computer programming andless for hand calculations).

2. Assume a point pressure q less than the maximum qb.

3. Read the corresponding displacement s from the (q - s) curve.

4. Assume that the load in the pile segment closest to the point (segment ri) is equal to thepoint load.

5. Compute the compression of the segment n under that load by

AAs =

where, Qp = qpAb,

A = cross-sectional area of segment,

E = modulus of elasticity of the pile material.

6. Calculate the settlement of the top of segment n by

1. Use the (i - s) curves to read the friction in on segment n, at displacement sn.

8. Calculate the load in pile segment (n - 1) by:


where Azn = length of segment n,

dn = average diameter of pile in segment n (applies to tapered piles).9. Do 4 through 8 up to the top segment. The load and displacement at the top of the pile

provide one point on the load-settlement curve.10. Repeat 1 through 9 for the other assumed values of the point pressure, q .

E may also be obtained from the relationship established between Es and the field tests suchas SPT, CPT and PMT. It may be noted here that the accuracy of the results obtained depends uponthe accuracy with which the values of Es simulate the field conditions.

Example 15.26A concrete pile of section 30 x 30 cm is driven into medium dense sand with the water table atground level. The depth of embedment of the pile is 18m. Static cone penetration test conducted atthe site gives an average value qc = 50 kg/cm2. Determine the load transfer curves and then calculatethe settlement. The modulus of elasticity of the pile material E is 21 x 104 kg/cm2 (Fig. Ex. 15.26).

Solution

It is first necessary to draw the (q - s) and (T- s) curves (see section 15.29). The curves can beconstructed by determining the ratios of q /s and t/s from Eqs (15.77) and (15.76) respectively.

Qp _ 3.125£s ~ 2R

6 m

6 m

6 m

\

f

f

f

3

Q

',

Q

'3

l

2

l

+ Sf

tr,1

T 53

tr 51 2 L-

f"-L „

Sand-= 18m

Square pile 30 x 30 cmEp = 21 x 104 kg/cm2

50

N 40S

"&> 30j*^ 20

10

0i

0.6

0.5

0.4rs

| 0.3M

^(-0.2

0.1

0

= 50kPa

^(max)=10.96 mmI I -

4 8 12Settlement, s, mm

ax = 0.55 kg/cm2

(r - s) curve

s(max)=1.71 mm

i i i i i0 0.5 1.0 1.5

Settlement, s, mm2.0

Figure Ex. 15.26

688 Chapter 15

T _ 0.22E

s ~ 2R

where R = radius or width of pileThe value of Es for a driven pile may be determined from Eq. (15.78).

Es = 3(36 + 2.2 qc) kg/cm2

= 3(36 + 2.2 x 50) = 438 kg/cm2

qp 3.125x438 ,Now — - = - = 45.62 kg/cnr

5 2x15

r 0.22x438 ,— = - = 3.21 kg/cnr^ 2x0.15

To construct the (q - s) and (T- s) curves, we have to know the maximum values of gp/s and T.

Given qb - q} = 50 kg/cm2 - the maximum value.

From Table 15.4 rmax = 0.01 1 qr = 0.01 1 x 50 = 0.55 kg/cm2

Now the theoretical maximum settlement s for ^(max) = 50 kg/cm2 is

=1-096 cm = 10.96 mm

The curve (q } - s) may be drawn as shown in Fig. Ex. 15.26. Similarly for T(max) = 0.55 kg/cm2

s = — — = 0.171 cm = 1.71 mm.(max) 3.21

Now the (T- s) curve can be constructed as shown in Fig. Ex. 15.26.

Calculation of pile settlement

The various steps in the calculations are

1. Divide the pile length 18 m into three equal parts of 6 m each.

2. To start with assume a base pressure q - 5 kg/cm2.

3. From the (q } - s} curve s, = 0.12 cm for q = 5 kg/cm2.

4. Assume that a load Q{ = Q = 5 x 900 = 4500 kg acts axially on segment 1 .

5. Now the compression As, of segment 1 is

. <2)AL 4500x600 .....As, = — — = - - = 0.014 cm1 AE, 30x30x21xl0 4

6. Settlement of the top of segment 1 is

s2 = S, + As, = 0.12 + 0.014 = 0.134 cm.7. Now from (r- s) curve Fig. Ex. 15.26, i- 0.43 kg/cm2.

8. The pile load in segment 2 is<22 = 4 x 30 x 600 x 0.43 + 4500 = 30,960 + 4500 = 35,460 kg

9. Now the compression of segment 2 is

<22AL 35,460x600As? = — - = - = 0.1 13 cm

" AEp 900x21xl04

10. Settlement of the top of segment 2 is


S2 = s2 + As2 = 0.134 + 0.113 = 0.247 cm.

11. Now from (T - s) curve, Fig. Ex. 15.26 T = 0.55 kg/cm2 for s3 = 0.247 cm. This is themaximum shear stress.

12. Now the pile load in segment 3 is

<23 = 4 x 30 x 600 x 0.55 + 35,460 = 39,600 + 35,460 = 75,060 kg

13. The compression of segment 3 is

_ Q3AL_ 75,060x600

AE 900x21xl04 = 0.238 cm

14. Settlement of top of segment 3 is

S4 = st = s2+&s2 = 0.247 + 0.238 = 0.485 cm.

15. Now from (T- s) curve, Tmax = 0.55 kg/cm2 for s > 0.17 cm.16. The pile load at the top of segment 3 is

QT = 4 x 30 x 600 x 0.55 + 75,060 = 39,600 + 75,060 = 114,660 kg= 115 tones (metric)

The total settlement st = 0.485 cm = 5 mm.

Total pile load QT = 115 tones.This yields one point on the load settlement curve for the pile. Other points can be obtainedin the same way by assuming different values for the base pressure q in Step 2 above. Foraccurate results, the pile should be divided into smaller segments.

15.30 SETTLEMENT OF PILE GROUPS IN COHESIVE SOILSThe total settlements of pile groups may be calculated by making use of consolidation settlementequations. The problem involves evaluating the increase in stress A/? beneath a pile group when thegroup is subjected to a vertical load Q . The computation of stresses depends on the type of soilthrough which the pile passes. The methods of computing the stresses are explained below:

Fictitiousfooting

\ / \ \

'A' iJJ.i_LLi

Weakerlayer

(a) (b) (c)

Figure 15.31 Settlement of pile groups in clay soils

690 Chapter 15

1. The soil in the first group given in Fig. 15.31 (a) is homogeneous clay. The load Q isassumed to act on a fictitious footing at a depth 2/3L from the surface and distributed overthe sectional area of the group. The load on the pile group acting at this level is assumed tospread out at a 2 Vert : 1 Horiz slope. The stress A/? at any depth z below the fictitiousfooting may be found as explained in Chapter 6.

2. In the second group given in (b) of the figure, the pile passes through a very weak layer ofdepth Lj and the lower portion of length L2 is embedded in a strong layer. In this case, theload Q is assumed to act at a depth equal to 2/3 L7 below the surface of the strong layer and

8 ^

spreads at a 2 : 1 slope as before.3. In the third case shown in (c) of the figure, the piles are point bearing piles. The load in this

case is assumed to act at the level of the firm stratum and spreads out at a 2 : 1 slope.

15.31 ALLOWABLE LOADS ON GROUPS OF PILESThe basic criterion governing the design of a pile foundation should be the same as that of a shallowfoundation, that is, the settlement of the foundation must not exceed some permissible value. Thepermissible values of settlements assumed for shallow foundations in Chapter 13 are alsoapplicable to pile foundations. The allowable load on a group of piles should be the least of thevalues computed on the basis of the following two criteria.

1. Shear failure,2. Settlement.

Procedures have been given in earlier chapters as to how to compute the allowable loads onthe basis of a shear failure criterion. The settlement of pile groups should not exceed thepermissible limits under these loads.

Example 15.27It is required to construct a pile foundation comprised of 20 piles arranged in 5 columns at distancesof 90 cm center to center. The diameter and lengths of the piles are 30 cm and 9 m respectively. Thebottom of the pile cap is located at a depth of 2.0 m from the ground surface. The details of the soilproperties etc. are as given below with reference to ground level as the datum. The water table wasfound at a depth of 4 m from ground level.

Soil properties

0

2

4

12

14

17

2

4

12

14

17

-

Silt, saturated, 7= 16 kN/m3

Clay, saturated, 7 = 19.2 kN/m3

Clay, saturated, 7= 19.2 kN/m3, qu = 120 kN/m2, eQ

Clay, 7= 18.24 kN/m3, qu = 90 kN/m2, eQ = 1.08, Cc

Clay, 7= 20 kN/m3, qu = 180 kN/m2, eQ = 0.70, Cc =

Rocky stratum

= 0.80,

= 0.34.

0.2

Cc = 0.23

Compute the consolidation settlement of the pile foundation if the total load imposed on thefoundation is 2500 kN.


Solution

Assume that the total load 2500 kN acts at a depth (2/3 )L = (2/3) x 9 = 6 m from the bottom of thepile cap on a fictitious footing as shown in Fig. 15. 31 (a). This fictitious footing is now at a depth of8 m below ground level. The size of the footing is 3.9 x 3.0 m. Now three layers are assumed tocontribute to the settlement of the foundation. They are: Layer 1 — from 8m to 12m (= 4m thick)below ground level; Layer 2 — from 1 2 m t o l 4 m = 2m thick; Layer 3 — from 14 m to 17 m = 3 mthick. The increase in pressure due to the load on the fictitious footing at the centers of each layer iscomputed on the assumption that the load is spread at an angle of 2 vertical to 1 horizontal[Fig. 15.31(a)] starting from the edges of the fictitious footing. The settlement is computed bymaking use of the equation

C p + Ac

where po = the effective overburden pressure at the middle of each layer,

A/7 = the increase in pressure at the middle of each layer

Computation of pa

For Layer 1, pg = 2x 16 + 2x 19.2 + (10 -4) (19.2 -9.81) = 126.74 kN/m2

For Layer 2, po = 126.74 + 2(19.2-9.81)+ lx (18.24-9.81) = 153.95 kN/m2

ForLayer3, po = 153.95 + 1(1 8.24 -9.81) + 1.5 x (20.0 -9.81) - 177.67 kN/m2

Computation of Ap

For Layer 1

Area at 2 m depth below fictitious footing = (3.9 + 2) x (3 + 2) = 29.5 m2

2500Ap = — — = 84.75 kN/m2

For Layer 2

Area at 5 m depth below fictitious footing = (3.9 + 5) x (3 + 5) = 71.2 m2

2500A/7 = — — = 35.1 kN/m2

For Layer 3

Area at 7.5 m below fictitious footing = (3.9 + 7.5) x (3 + 7.5) = 1 19.7 m2

= 20.9 kN/m2

119.7

Settlement computation

, c 4x023 126.74 + 84.75Layer 1 S, = - log - = 0.1 13m

l 1 + 0.80 126.74

2x0.34, 153.95 + 35.1 nMnLayer 2 59 = - log - = 0.029 m3 2 1 + 1.08 153.95

692 Chapter 15

T _ c 3x0.2, 177.67 + 20.9 n m _Layer 3 S3 = -—— log = 0.017 m

1 + 0.7

Total = 0.159m- 16cm.

177.67

15.32 NEGATIVE FRICTIONFigure 15.32(a) shows a single pile and (b) a group of piles passing through a recently constructedcohesive soil fill. The soil below the fill had completely consolidated under its overburden pressure.

When the fill starts consolidating under its own overburden pressure, it develops a drag on thesurface of the pile. This drag on the surface of the pile is called 'negative friction'. Negative friction

e»1

Fill -H ' Negative- .

• • VJ:B-friction •'•'•

•: Natural £\ M ••& Frictional ;•.• Stiff Soil '..^'^^•'-*'- ffcicfanr'p •.*! resistance ;

' -•>•. ".N .. •••>••. ''fit ••.•:••

Point resistance

(a)

:

Fill

^ Natural.' stiff soil

i .V-V'l•;i 4*3^

Figure 15.32 Negative friction on piles


may develop if the fill material is loose cohesionless soil. Negative friction can also occur when fillis placed over peat or a soft clay stratum as shown in Fig. 15.32c. The superimposed loading onsuch compressible stratum causes heavy settlement of the fill with consequent drag on piles.

Negative friction may develop by lowering the ground water which increases the effectivestress causing consolidation of the soil with resultant settlement and friction forces beingdeveloped on the pile.

Negative friction must be allowed when considering the factor of safety on the ultimatecarrying capacity of a pile. The factor of safety, Fs, where negative friction is likely to occur may bewritten as

Ultimate carrying capacity of a single pile or group of pilesF =s Working load + Negative skin friction load

Computation of Negative Friction on a Single Pile

The magnitude of negative friction Fn for a single pile in a fill may be taken as (Fig. 15.32(a)).

(a) For cohesive soils

Fn = PLns. (15.80)

(b) For cohesionless soils

(15.81)

where Ln = length of piles in the compressible material,s = shear strength of cohesive soils in the fill,P = perimeter of pile,K = earth pressure coefficient normally lies between the active and the passive earth

pressure coefficients,S = angle of wall friction which may vary from (|)/2 to ()).

Negative Friction on Pile Groups

When a group of piles passes through a compressible fill, the negative friction, Fn , on the groupmay be found by any of the following methods [Fig. 15.32b].

(a) Fng = nFn (15.82)

(b) Fng = sLnPg + rLAg (15.83)

where n = number of piles in the group,y = unit weight of soil within the pile group to a depth Ln,

P = perimeter of pile group,A - sectional area of pile group within the perimeter P ,

o o

s = shear strength of soil along the perimeter of the group.Equation (15.82) gives the negative friction forces of the group as equal to the sum of the

friction forces of all the single piles.Eq. (15.83) assumes the possibility of block shear failure along the perimeter of the group

which includes the volume of the soil yLnA enclosed in the group. The maximum value obtainedfrom Eqs (15.82) or (15.83) should be used in the design.

When the fill is underlain by a compressible stratum as shown in Fig. 15.32(c), the totalnegative friction may be found as follows:

694 Chapter 15

Fn8 = "(Fnl + (15-84)

Fng = S\L\Pg + S2L2Pg + XlM* + Mg

- Pg(s}L{ + s2L2) + Ag(y]Ll + r2L2) (15.85)

where L} = depth of fill,L, = depth of compressible natural soil,

SP s? = shear strengths of the fill and compressible soils respectively,yp 72 = unit weights of fill and compressible soils respectively,

Fnl = negative friction of a single pile in the fill,Fn2 = negative friction of a single pile in the compressible soil.

The maximum value of the negative friction obtained from Eqs. (15.84) or (15.85) should beused for the design of pile groups.

Example 15.28A square pile group similar to the one shown in Fig. 15.27 passes through a recently constructedfill. The depth of fill Ln - 3 m. The diameter of the pile is 30 cm and the piles are spaced 90 cmcenter to center. If the soil is cohesive with qu = 60 kN/m2, and y= 15 kN/m3, compute the negativefrictional load on the pile group.

Solution

The negative frictional load on the group is the maximum of [(Eqs (15.82) and (15.83)]

(a) F = nF , and (b) F = sL P + vL A ,^ > ng n' v ' ng n g ' n #'

where P = 4 x 3 = 12 m, Ag = 3 x 3 = 9 m2, cu = 60/2 = 30 kN/rn2

(a) Fn = 9 x 3.14x0.3x3x30-763 kN

(b) Fng = 30x3x12 + 1 5 x 3 x 9 -1485 kN

The negative frictional load on the group = 1485 kN.

15.33 UPLIFT CAPACITY OF A PILE GROUPThe uplift capacity of a pile group, when the vertical piles are arranged in a closely spaced groupsmay not be equal to the sum of the uplift resistances of the individual piles. This is because, atultimate load conditions, the block of soil enclosed by the pile group gets lifted. The manner inwhich the load is transferred from the pile to the soil is quite complex. A simplified way ofcalculating the uplift capacity of a pile group embedded in cohesionless soil is shown inFig. 15.33(a). A spread of load of 1 Horiz : 4 Vert from the pile group base to the ground surfacemay be taken as the volume of the soil to be lifted by the pile group (Tomlinson, 1977). Forsimplicity in calculation, the weight of the pile embedded in the ground is assumed to be equal tothat of the volume of soil it displaces. If the pile group is partly or fully submerged, the submergedweight of soil below the water table has to be taken.

In the case of cohesive soil, the uplift resistance of the block of soil in undrained shearenclosed by the pile group given in Fig. 15.33(b) has to be considered. The equation for the totaluplift capacity P u of the group may be expressed by

P,u=2L(L + B)cu + W (15.86)


t t I t t

Block of soillifted by piles

(a) Uplift of a group of closely-spaced piles in cohesionless soils

t t I t t

Block of soillifted by piles

Bx L-

(b) Uplift of a group of piles in cohesive soils

Figure 15.33 Uplift capacity of a pile group

where L = depth of the pile block

L and B = overall length and width of the pile groupcu = average undrained shear strength of soil around the sides of the groupW = combined weight of the block of soil enclosed by the pile group plus the weight of

the piles and the pile cap.A factor of safety of 2 may be used in both cases of piles in sand and clay.

The uplift efficiency E of a group of piles may be expressed as

(15.89)

where PUS = uplift capacity of a single pile

n = number of piles in the group

The efficiency E u varies with the method of installation of the piles, length and spacing andthe type of soil. The available data indicate that E increases with the spacing of piles. Meyerhof andAdams (1968) presented some data on uplift efficiency of groups of two and four model circular

696 Chapter 15

footings in clay. The results indicate that the uplift efficiency increases with the spacing of thefootings or bases and as the depth of embedment decreases, but decreases as the number of footingsor bases in the group increases. How far the footings would represent the piles is a debatable point.For uplift loading on pile groups in sand, there appears to be little data from full scale field tests.

15.34 PROBLEMS

15.1

15.2

A 45 cm diameter pipe pile of length 12m with closed end is driven into a cohesionless soilhaving 0 = 35°. The void ratio of the soil is 0.48 and Gs = 2.65. The water table is at theground surface. Estimate (a) the ultimate base load Qb, (b) the frictional load <2« and (c) theallowable load Qa with F , - 2.5.Use the Berezantsev method for estimating Qb. For estimating Q~ use Ks = 0.75 and5=20°.

~

Refer to Problem 15.1. Compute Qb by Meyerhof's method. Determine Q, using thecritical depth concept, and Qa with F? = 2.5. All the other data given in Prob. 15.1 remainthe same.

15.3 Estimate Qb by Vesic's method for the pile given in Prob. 15.1. Assume Ir=In = 60.Determine Qa for Fy = 2.5 and use (X obtained in Prob. 15.1.

15.4 For Problem 15.1, estimate the ultimate base resistance Qb by Janbu's method. DetermineQa with Fs = 2.5. Use Q/ obtained in Prob. 15.1. Use (// = 90°.

15.5 For Problem 1 5.1, estimate Qb, Q,, and Qa by Coyle and Castello method. All the data givenremain the same.

15.6 For problem 15.1, determine Qb, and Qa by Meyerhof's method using the relationshipbetween Ncor and 0 given in Fig 12.8.

,

18.

//&\ //A$*.

10

> m1

,

6

1

,

2.11

m

m

m\

1 '

XWs /5'Ws X*5CX X^C^v

cu = 30 kN/m2

0 = 0

y = 18.0 kN/m3

/ d = 40 cm

cu = 50 kN/m2

0 = 0

y = 18.5 kN/m3

cu= 150 kN/m2

0 = 0= 19.0kN/mJ

Figure Prob. 15.1 1


15.7 A concrete pile 40 cm in diameter is driven into homogeneous sand extending to a greatdepth. Estimate the ultimate load bearing capacity and the allowable load with F = 3.0 byCoyle and Castello's method. Given: L = 15 m, 0 = 36°, y= 18.5 kN/m3.

15.8 Refer to Prob. 15.7. Estimate the allowable load by Meyerhof s method using therelationship between 0 and Ncor given in Fig. 12.8.

15.9 A concrete pile of 15 in. diameter, 40 ft long is driven into a homogeneous's stratum of claywith the water table at ground level. The clay is of medium stiff consistency with theundrained shear strength cu = 600 lb/ft2. Compute Qb by Skempton's method and (^by theor-method. Determine Qa for Fs = 2.5.

15.10 Refer to Prob 15.9. Compute Q,by the A-method. Determine Qa by using Qb computed inProb. 15.9. Assume ysat = 120 lb/ft2.

15.11 A pile of 40 cm diameter and 18.5 m long passes through two layers of clay and isembedded in a third layer. Fig. Prob. 15.11 gives the details of the soil system. Compute Q,by the a - method and Qb by Skempton's method. Determine Qa for F5 = 2.5.

15.12 A concrete pile of size 16 x 16 in. is driven into a homogeneous clay soil of mediumconsistency. The water table is at ground level. The undrained shear strength of the soil is500 lb/ft2. Determine the length of pile required to carry a safe load of 50 kips with FS = 3.Use the a - method.

15.13 Refer to Prob. 15.12. Compute the required length of pile by the X-method. All the otherdata remain the same. Assume ysat =120 lb/ft3.

15.14 A concrete pile 50 cm in diameter is driven into a homogeneous mass of cohesionless soil.The pile is required to carry a safe load of 700 kN. A static cone penetration test conductedat the site gave an average value of qc = 35 kg/cm2 along the pile and 60 kg/cm2 below thebase of the pile. Compute the length of the pile with Fs=3.

15.15 Refer to Problem 15.14. If the length of the pile driven is restricted to 12 m, estimate theultimate load Qu and safe load Qa with Fs = 3. All the other data remain the same.

15.16 A reinforced concrete pile 20 in. in diameter penetrates 40 ft into a stratum of clay and restson a medium dense sand stratum. Estimate the ultimate load.

Given: for sand- 0= 35°, ysat = 120 lb/ft3

for clay ysat = 119 lb/ft3, cu = 800 lb/ft2.

Use (a) the cc-method for computing the frictional load, (b) Meyerhof's method forestimating Qb. The water table is at ground level.

15.17 A ten-story building is to be constructed at a site where the water table is close to theground surface. The foundation of the building will be supported on 30 cm diameter pipepiles. The bottom of the pile cap will be at a depth of 1.0 m below ground level. The soilinvestigation at the site and laboratory tests have provided the saturated unit weights, theshear strength values under undrained conditions (average), the corrected SPT values, andthe soil profile of the soil to a depth of about 40 m. The soil profile and the other details aregiven below.

Depth (m)

From

06.022

30

To

6

22

30

40

Soil

Sand

Med. stiff clay

sand

stiff clay

v N'sat cor

kN/m 3

19 18

18

19.6 25

18.5

</>° c (average)

kN/m 2

33°

60

35°

75

698 Chapter 15

Determine the ultimate bearing capacity of a single pile for lengths of (a) 15m, and (b)25 m below the bottom of the cap.

Use a = 0.50 and Ks = 1.2. Assume S = 0.8 0.

15.18 For a pile designed for an allowable load of 400 kN driven by a steam hammer (singleacting) with a rated energy of 2070 kN-cm, what is the approximate terminal set of the pileusing the ENR formula?

15.19 A reinforced concrete pile of 40 cm diameter and 25 m long is driven through mediumdense sand to a final set of 2.5 mm, using a 40 kN single - acting hammer with a stroke of150 cm. Determine the ultimate driving resistance of the pile if it is fitted with a helmet,plastic dolly and 50 mm packing on the top of the pile. The weight of the helmet, withdolly is 4.5 kN. The other particulars are: weight of pile = 85 kN, weight of hammer 35 kN;pile hammer efficiency TJ h = 0.85; the coefficient restitution Cr = 0.45. Use Hiley'sformula. The sum of elastic compression C = cl + c2 + c3 = 20.1 mm.

15.20 A reinforced concrete pile 45 ft long and 20 in. in diameter is driven into a stratum ofhomogeneous saturated clay having cu = 800 lb/ft2. Determine (a) the ultimate loadcapacity and the allowable load with Fs = 3; (b) the pullout capacity and the allowablepullout load with Fs = 3. Use the a-method for estimating the compression load.

15.21 Refer to Prob. 15.20. If the pile is driven to medium dense sand, estimate (a) the ultimatecompression load and the allowable load with FS = 3, and (b) the pullout capacity and theallowable pullout load with Fs = 3. Use the Coyle and Castello method for computing Qb

and Q,. The other data available are: 0 = 36°, and 7 = 115 lb/ft3. Assume the water table isat a great depth.

15.22 A group of nine friction piles arranged in a square pattern is to be proportioned in a depositof medium stiff clay. Assuming that the piles are 30 cm diameter and 10m long, find theoptimum spacing for the piles. Assume a = 0.8 and cu - 50 kN/m2.

15.23 A group of 9 piles with 3 in a row was driven into sand at a site. The diameter and length ofthe piles are 30 cm and 12m respectively. The properties of the soil are: 0= 30, e - 0.7, andGs = 2.64.If the spacing of the piles is 90 cm, compute the allowable load on the pile group on thebasis of shear failure for Fs = 2.0 with respect to skin resistance, and Fs = 2.5 with respectto base resistance. For 0 = 30°, assume N = 22.5 and N = 19.7. The water table is atground level.

15.24 Nine RCC piles of diameter 30 cm each are driven in a square pattern at 90 cm center tocenter to a depth of 12 m into a stratum of loose to medium dense sand. The bottom of thepile cap embedding all the piles rests at a depth of 1.5 m below the ground surface. At adepth of 15 m lies a clay stratum of thickness 3 m and below which lies sandy strata. Theliquid limit of the clay is 45%. The saturated unit weights of sand and clay are 18.5 kN/m3

and 19.5 kN/m3 respectively. The initial void ratio of the clay is 0.65. Calculate theconsolidation settlement of the pile group under the allowable load. The allowable loadQ = 120 kN.^•a

15.25 A square pile group consisting of 16 piles of 40 cm diameter passes through two layers ofcompressible soils as shown in Fig. 15.32(c). The thicknesses of the layers are : Lj = 2.5 mand L2 = 3 m. The piles are spaced at 100 cm center to center. The properties of the fillmaterial are: top fill cu = 25 kN/m2; the bottom fill (peat), cu = 30 kN/m2. Assume7= 14 kN/m3 for both the fill materials. Compute the negative frictional load on the pilegroup.

Ok_ch15-Deep Foundation I-pile Foundation

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Transcript of Ok_ch15-Deep Foundation I-pile Foundation